RD Sharma Class 12 Exercise 18.25 Indefinite integrals Solutions Maths

RD Sharma Class 12 Exercise 18.25 Indefinite integrals Solutions Maths - Download PDF Free Online

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Class 12 RD Sharma chapter 18 exercise 18.25 solution provides the best material for maths for CBSE students. It has been highly recommended by students, teachers as well as experts for the thorough practice to perform well in the board exams. The RD Sharma class 12th exercise 18.25 will give you the best solution so that you will not find any need to refer to any resource for help while solving maths.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.25
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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.25

Indefinite Integrals Exercise 18.25 Question 1

Answer: $x \sin x + \cos x + c$
Hint: $\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x$
Given: $I = \int x \cos x d x$
Solution:Using integration by parts

I = x \int \cos x d x - \int \frac{d}{d x} x \int \cos x d x

We have,

\int \sin x d x = - \cos x, \int \cos x d x = \sin x

\begin{aligned} = x \times \sin x - \int \sin x d x \\ = x \sin x + \cos x + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 2

Answer: $x \log (x + 1) - x + \log (x + 1) + c$
Hint: We use integration by parts,

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: Let $I = \int \log (x + 1) d x$
Solution: $I = \int 1 \times \log (x + 1) d x$
$I = \log (x + 1) \int 1 d x - \int \frac{d}{d x} \log (x + 1) \int 1 d x$
We know that,

\int 1 d x = x & \frac{d}{d x} \log x = \frac{1}{x}

= \log (x + 1) \times x - \int \frac{1}{x + 1} \times x

\begin{aligned} = \frac{x}{x + 1} = 1 - \frac{1}{x + 1} \\ = x \log (x + 1) - \int (1 - \frac{1}{x + 1}) d x \\ = x \log (x + 1) - x + \log (x + 1) + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 3

Answer: $\frac{x^{4}}{4} \log x - \frac{x^{4}}{16} + c$
Hint: Use Integration by parts

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: $I = \int x^{3} \log x d x$
Solution: On using Integration by parts

I = \log x \int x^{3} d x - \int \frac{d}{d x} \log x \int x^{3} d x

We have,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} & \frac{d}{d x} \log x = \frac{1}{x}

\begin{aligned} = \log x \times \frac{x^{4}}{4} - \int \frac{1}{x} \times \frac{x^{4}}{4} d x \\ = \log x \times \frac{x^{4}}{4} - \frac{1}{4} \int x^{3} d x \\ = \frac{x^{4}}{4} \log x - \frac{1}{4} \times \frac{x^{4}}{4} \\ = \frac{x^{4}}{4} \log x - \frac{x^{4}}{16} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 4

Answer: $x e^{x} - e^{x} + c$
Hint: Using integration by parts
$\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x$
Given: $I = \int x e^{x} d x$
Solution: $I = x \int e^{x} d x - \int \frac{d}{d x} x \int e^{x} d x$
We have,

\int e^{x} d x = e^{x} & \frac{d}{d x} x = 1

\begin{aligned} = x e^{x} - \int e^{x} d x \\ = x e^{x} - e^{x} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 5

Answer: $(\frac{x}{2} - \frac{1}{4}) e^{2 x} + c$
Hint:Use integration by parts,

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: Let, $I = \int x e^{2 x} d x$
Solution: On using integration by parts

I = x \int e^{2 x} d x - \int \frac{d}{d x} x \int e^{2 x} d x

We Know that,

\int e^{r x} d x = \frac{e^{r x}}{n} & \frac{d}{d x} x = 1

\begin{aligned} = \frac{x e^{2 x}}{2} - \int \frac{e^{2 x}}{2} d x \\ = \frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4} + c \\ I = (\frac{x}{2} - \frac{1}{4}) e^{2 x} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 6

Answer: $- e^{- x} (x^{2} + 2 x + 2) + c$
Hint: Integration by parts says that:

\begin{aligned} \int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x \\ u = x^{2}, \frac{d u}{d x} = 2 x \\ \frac{d v}{d x} = - e^{- x}, v = e^{- x} \\ \int x^{2} e^{- x} d x = - x^{2} e^{- x} - \int - 2 x e^{- 2 x} d x \end{aligned}

Given: $I = \int x^{2} e^{- x} d x$
Solution: $\int - 2 x e^{- 2 x} d x$
$\begin{aligned} u = 2 x, \frac{d u}{d x} = 2 \\ \frac{d v}{d x} = - e^{- x}, v = e^{- x} \\ = \int - 2 x e^{- 2 x} d x \\ = 2 x e^{- x} - \int 2 e^{- x} d x \\ = 2 x e^{- x} + 2 e^{- x} \\ = \int x^{2} e^{- x} d x \\ = - x^{2} e^{- x} - (2 x e^{- x} + 2 e^{- x}) \\ = - x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x} + c \\ = - e^{- x} (x^{2} + 2 x + 2) + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 7

Answer: $x^{2} \sin x + 2 x \cos x - 2 \sin x + c$
Hint: Use integration by parts

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: $I = \int x^{2} \cos x d x$
Solution: $I = \int x^{2} \cos x d x$
$= x^{2} \int \cos x d x - \int \frac{d}{d x} x^{2} \int (\cos x d x) d x$
By integrating w.r.t x

= x^{2} \sin x - \int 2 x \sin x d x

So we get,

= x^{2} \sin x - 2 [\int x \sin x d x]

Now apply by parts method

\begin{aligned} = x^{2} \sin x - 2 [\int x \sin x d x] \\ = x^{2} \sin x - 2 [x \int \sin x d x - \int (\frac{d}{d x} x \int \sin x d x) d x] \\ = x^{2} \sin x - 2 [x (- \cos x) - \int 1 (- \cos x) d x] \\ = x^{2} \sin x - 2 (- x \cos x + \sin x) + c \end{aligned}

On further simplification,

= x^{2} \sin x + 2 x \cos x - 2 \sin x + c

Indefinite Integrals Exercise 18.25 Question 8

Answer: $\frac{x^{2} \sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4} + c$
Hint:Use integration by parts

\begin{aligned} \int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x \\ Let u = x^{2} & v = \cos 2 x \end{aligned}

Given: $I = \int x^{2} \cos 2 x d x$
Solution: $I = \int x^{2} \cos 2 x d x$
$\begin{aligned} = x^{2} \int \cos 2 x d x - \int \frac{d}{d x} x^{2} \int (\cos 2 x d x) d x \\ = x^{2} (\frac{\sin 2 x}{2}) - \int 2 x (\frac{\sin 2 x}{2}) d x \\ = \frac{x^{2} \sin 2 x}{2} - [\int x \sin 2 x d x] \\ = \frac{x^{2} \sin 2 x}{2} - [x \int \sin 2 x d x - \int (\frac{d x}{d x} \int \sin 2 x d x) d x] \end{aligned}$
Integrate on by parts applying once again

\begin{aligned} = \frac{x^{2} \sin 2 x}{2} - [x (\frac{- \cos 2 x}{2}) - \int (\frac{- \cos 2 x}{2}) d x] \\ = \frac{x^{2} \sin 2 x}{2} - [\frac{- x \cos 2 x}{2} + \int \frac{\cos 2 x}{2} d x] \\ = \frac{x^{2} \sin 2 x}{2} - [\frac{- x \cos 2 x}{2} + \frac{\sin 2 x}{4}] + c \\ = \frac{x^{2} \sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 9

Answer: $\frac{- 1}{2} x \cos 2 x + \frac{1}{4} \sin 2 x + c$
Hint: Use integrate on by parts

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: $I = \int x \sin 2 x d x$
Solution: $I = \int x \sin 2 x d x$
$\Rightarrow d (\cos 2 x) = - 2 \sin 2 x d x$
We can integrate by parts in this way

\begin{aligned} = \int x \sin 2 x d x \\ = \frac{- 1}{2} \int x d (\cos 2 x) \\ = \frac{- 1}{2} x \cos 2 x + \frac{1}{2} \int \cos 2 x d x \\ = \frac{- 1}{2} x \cos 2 x + \frac{1}{4} \sin 2 x + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 10

Answer: $\log x {\log (\log x) - 1} + c$
Hint:Take $\log x = t So, \frac{1}{x} d x = d t$
Then,

\int \frac{\log (\log x)}{x} d x = \int \log t d t = \int 1 \cdot \log t d t

Consider the 1st function as log t and 2nd function
Given: $I = \int \frac{\log (\log x)}{x} d x$
Solution:
Let $\log x = t So, \frac{1}{x} d x = d t$
$\begin{aligned} I = \int 1 \cdot \log t d t \\ = \log t \int 1 d t - \int (\frac{d \log t}{d t} \cdot \int 1 d t) d t \end{aligned}$
By integrating w.r.t ‘t’

= t \log t - \int \frac{1}{t} \cdot t d t

Again by integrating the second term we get

= t \log t - t + c

Replacing t with logx

= \log x \cdot \log (\log x) - \log x + c

By taking log x as common

= \log x {\log (\log x) - 1} + c

Indefinite Integrals Exercise 18.25 Question 11

Answer: $x^{2} \sin x + 2 x \cos x - 2 \sin x + c$
Hint: Use integration by parts

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: Let

I = \int x^{2} \cos x d x

Solution: $I = \int x^{2} \cos x d x$
$\begin{aligned} = x^{2} \int \cos x d x - \int (\frac{d}{d x} x^{2} \int \cos x d x) d x \\ = x^{2} \sin x - 2 [\int x \sin x d x] \end{aligned}$
Now apply by part method again

\begin{aligned} = x^{2} \sin x - 2 \int x \sin x d x \\ = x^{2} \sin x - 2 [x \int \sin x d x - \int (\frac{d}{d x} x \cdot \int \sin x d x) d x] \\ = x^{2} \sin x - 2 [x (- \cos x) - \int 1 . (- \cos x) d x] \\ = x^{2} \sin x - 2 (- x \cos x + \sin x) + c \end{aligned}

On further simplification,

= x^{2} \sin x + 2 x \cos x - 2 \sin x + c

Indefinite Integrals Exercise 18.25 Question 12

Answer: $- x \cot x + \ln | \sin x | + c$
Hint: Here, using Integration by parts

\int a b d x = a \int b d x - \int [\frac{d}{d x} a \int b d x] d x

Given: $I = \int x {cosec}^{2} x d x$
Solution: $I = \int x {cosec}^{2} x d x$
$I = x \int {cosec}^{2} x - \int [\frac{d x}{d x} \cdot \int \cos e c^{2} x d x] d x$
$\begin{aligned} = x (- \cot x) - \int 1 (- \cot x) d x \\ = - x \cot x + \int \cot x d x \\ = - x \cot x + \ln | \sin x | + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 13

Answer: $\frac{x^{2}}{4} + \frac{x \sin (2 x)}{4} + \frac{\cos (2 x)}{8} + c$
Hint:Using integration by parts

\int a b d x = a \int b d x - \int [\frac{d}{d x} a \int b d x] d x

Given: $I = \int x \cos^{2} x d x$
Solution: $\int x \cos^{2} x d x = \int x \cdot \frac{1 + \cos 2 x}{2} d x$
Break this integral apart now

\begin{aligned} = \int x \cdot (\frac{1 + \cos 2 x}{2}) d x \\ = \int \frac{x}{2} d x + \int \frac{x \cos (2 x)}{2} d x \end{aligned}

Evaluate the integral part by part

\begin{aligned} \int \frac{x}{2} d x = \frac{x^{2}}{4} + c \\ \int \frac{x \cos (2 x)}{2} d x \end{aligned}

Integration by parts
Let

u = \frac{\cos (2 x)}{2}

\int u d x = \frac{\sin 2 x}{4}

\int \frac{x \cos (2 x)}{2} d x = \frac{x \cdot \sin (2 x)}{4} - \int \frac{\sin 2 x}{4} d x

Now

\int \frac{- \sin (2 x)}{4} d x = \frac{\cos (2 x)}{8} + c

Combining all,

\int x \cos^{2} (x) d x = \frac{x^{2}}{4} + \frac{x \cdot \sin (2 x)}{4} + \frac{\cos (2 x)}{8} + c

Indefinite Integrals Exercise 18.25 Question 14

Answer: $\frac{x^{n + 1}}{n + 1} [\log x - \frac{1}{n + 1}] + c$
Hint: $\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x$
Where, $u = \log x, d v = x^{n} d x$
Given: Let $I = \int x^{n} \log x d x$
Solution:
$\begin{aligned} d u = \frac{1}{x} (v = \frac{x^{n + 1}}{n + 1}) \\ = \frac{x^{n + 1}}{n + 1} \log x - \int \frac{x^{n + 1}}{n + 1} (\frac{1}{x}) \\ = \frac{x^{n + 1}}{n + 1} \log x - \frac{1}{n + 1} \int x^{n} d x \\ = \frac{x^{n + 1}}{n + 1} \log x - \frac{x^{n + 1}}{(n + 1) (x + 1)} + c \\ = \frac{x^{n + 1}}{n + 1} [\log x - \frac{1}{n + 1}] + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 15

Answer: $\frac{x^{- n + 1}}{1 - n} \log x - \frac{x^{- n + 1}}{(1 - n)^{2}} + c$
Hint: Use integration by parts

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: $Let I = \int \frac{\log x}{x^{n}} d x$
Solution: $I = \int \frac{\log x}{x^{n}} d x$
$\begin{aligned} = \int x^{- n} \log x d x \\ = \log x \int x^{- n} d x - \int (\frac{d}{d x} \log x \int x^{- n} d x) d x \\ = \log x (\frac{x^{- n + 1}}{- n + 1}) - \int \frac{1}{x} \cdot \frac{x^{- n + 1}}{- n + 1} d x \\ = \frac{x^{- n + 1} \log x}{1 - n} - \frac{1}{1 - n} \int \frac{x^{- n} x}{x} d x \end{aligned}$
Again by integrating the second part

= \frac{x^{- n + 1} \log x}{1 - n} - \frac{1}{1 - n} \times \frac{x^{- n + 1}}{- n + 1} + c

So we get

$= \frac{x^{- n + 1}}{1 - n} \log x - \frac{x^{- n + 1}}{(1 - n)^{2}} + c$

Indefinite Integrals Exercise 18.25 Question 16

Answer: $\frac{x^{3}}{6} - \frac{x^{2} \sin 2 x}{4} - \frac{x \cos 2 x}{4} + \frac{\sin 2 x}{8} + c$
Hint: Use the formula

1) \sin^{2} x = \frac{1 - \cos 2 x}{2}

2) \int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given: $I = \int x^{2} \sin^{2} x d x$
Solution: $I = \int x^{2} \sin^{2} x d x$
$\begin{aligned} \int x^{2} (\frac{1 - \cos 2 x}{2}) d x = \int \frac{x^{2}}{2} - \frac{x^{2} \cos 2 x}{2} d x \\ = \int \frac{x^{2}}{2} d x - \int \frac{x^{2} \cos 2 x}{2} d x \\ = \frac{x^{3}}{3 \times 2} - \frac{1}{2} \int x^{2} \cos 2 x d x \\ = \frac{x^{3}}{6} - \frac{1}{2} (x^{2} \int \cos 2 x d x - \int (\frac{d}{d x} x^{2} \int \cos 2 x d x) d x) \\ = \frac{x^{3}}{6} - \frac{1}{2} (x^{2} \frac{\sin 2 x}{2} - \int 2 x \frac{\sin 2 x}{2} d x) \\ = \frac{x^{3}}{6} - \frac{1}{2} (x^{2} \frac{\sin 2 x}{2} - \int x \sin 2 x d x) \end{aligned}$
Now by integrating the second part we have,

\begin{aligned} = \frac{x^{3}}{6} - \frac{1}{2} (x^{2} \frac{\sin 2 x}{2} - [x \frac{- \cos 2 x}{2} - \int 1 \frac{- \cos 2 x}{2} d x]) \\ = \frac{x^{3}}{6} - \frac{1}{2} (x^{2} \frac{\sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4}) + c \\ = \frac{x^{3}}{6} - \frac{x^{2} \sin 2 x}{4} - \frac{x \cos 2 x}{4} + \frac{\sin 2 x}{8} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 17

Answer: $e^{x^{2}} (x^{2} - 1) + c$
Hint:Take, $x^{2} = t$
So we get $2 x d x = d t$
Given: Let $I = \int 2 x^{3} e^{x^{2}} d x$
$= \int 2 x x^{2} e^{x^{2}} d x = \int t . e^{t} d t$
By integrating w.r.t ‘t’ taking the first function as t and second function as

e^{t}

$\begin{aligned} = \int t . e^{t} d t = t \int e^{t} d t - \int (\frac{d}{d t} t \int e^{t} d t) d t \\ = t e^{t} - e^{t} + c \end{aligned}$
Now by replacing t with

x^{2}

x^{2} e^{x^{2}} - e^{x^{2}} + c

By taking

e^{x^{2}}

as common

= e^{x^{2}} (x^{2} - 1) + c

Indefinite Integrals Exercise 18.25 Question 18

Answer: $\frac{1}{2} x^{2} \sin x^{2} + \frac{1}{2} \cos x^{2} + c$
Hint: Take $x^{2} = t$
So we get

2 x d x = d t

x d x = \frac{d t}{2}

Given: Let $I = \int x^{3} \cos x^{2} d x$
Solution: $I = \int x^{3} \cos x^{2} d x$
We can write it as

\begin{aligned} \int x x^{2} \cos x^{2} d x \\ = \frac{1}{2} \int t \cos t d t = \frac{1}{2} (t \int \cos t d t - \int [\frac{d t}{d t} \int \cos t d t] d t) \\ = \frac{1}{2} (t \sin t - \int \sin t d t) \end{aligned}

Now by integrating the second part

= \frac{1}{2} (t \sin t + \cos t) + c

Substituting the value of t as

x^{2}

= \frac{1}{2} x^{2} \sin x^{2} + \frac{1}{2} \cos x^{2} + c

Indefinite Integrals Exercise 18.25 Question 19

Answer: $\frac{- x \cos 2 x}{4} + \frac{\sin 2 x}{8} + c$
Hint: We know that,
Sin 2x = 2 sinxcosx, it can be written as

\int x \sin x \cos x d x = \frac{1}{2} \int x \sin 2 x d x

and consider first
function as x and second function as sin2x
Given: Let $I = \int x \sin x \cos x d x$
Solution: $\frac{1}{2} \int x \sin 2 x d x$
$\begin{aligned} = \frac{1}{2} (x \int \sin 2 x d x - \int [\frac{d}{d x} x \int \sin 2 x d x] d x) \\ = \frac{1}{2} (x \frac{- \cos 2 x}{2} + \int \frac{\cos 2 x}{2} d x) \end{aligned}$
Again by integrating second term

= \frac{1}{2} (\frac{- x \cos 2 x}{2} + \frac{\sin 2 x}{4}) + c

By further simplification,

$= \frac{- x \cos 2 x}{4} + \frac{\sin 2 x}{8} + c$

Indefinite Integrals Exercise 18.25 Question 20

Answer: $- \cos x \log (\cos x) + \cos x + c$
Hint: Consider the first function as log(cos x) and second function as sin x
Given: Let $I = \int \sin x \log (\cos x) d x$
Solution: $I = \int \sin x \log (\cos x) d x$
$\begin{aligned} = \log (\cos x) \int \sin x d x - \int (\frac{d}{d x} \log (\cos x) \int \sin x d x) d x \\ = - \cos x \log (\cos x) + \int \frac{- \sin x}{\cos x} \cos x d x \end{aligned}$
So we get,

= - \cos x \log (\cos x) - \int \sin x d x

Again by integrating the second term

= - \cos x \log (\cos x) + \cos x + c

Indefinite Integrals Exercise 18.25 Question 21

Answer:
$= \frac{x^{2}}{2} (\log x)^{2} - \frac{x^{2}}{2} \log x + \frac{x^{2}}{4} + c$
Hint: Taking

(\log x)^{2}

as first function and 1-> Second function and integration

\int u v d x = u \int v d x - \int \frac{d}{d x} u \int v d x

Given:
$\int x (\log x)^{2} d x$
Solution:
$I = \int x (\log x)^{2} d x$
Taking

(\log x)^{2}

as first function and x as Second function and integrate

\begin{aligned} I = (\log x)^{2} \int x d x - \int {[\frac{d}{d x} (\log x)^{2} \int x d x]} d x \\ = \frac{x^{2}}{2} (\log x)^{2} - [\int 2 \log x \frac{1}{x} \frac{x^{2}}{2} d x] \Rightarrow \frac{x^{2}}{2} (\log x)^{2} - \int x \log x d x \end{aligned}

Again integrating by parts, we obtain

\begin{aligned} I = \frac{x^{2}}{2} (\log x)^{2} - [\log x \int x d x - \int [\frac{d}{d x} \log x] \int x d x] d x \\ = \frac{x^{2}}{2} (\log x)^{2} - [\frac{x^{2}}{2} \log x + \int \frac{1}{x} \frac{x^{2}}{2} d x] \\ = \frac{x^{2}}{2} (\log x)^{2} - \frac{x^{2}}{2} \log x + \frac{1}{2} \int x d x \\ = \frac{x^{2}}{2} (\log x)^{2} - \frac{x^{2}}{2} \log x + \frac{x^{2}}{4} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 22

Answer: $2 e^{\sqrt{x}} (\sqrt{x} - 1) + c$
Hint: Put $e^{\sqrt{x}} = t$
Given: $\int e^{\sqrt{x}} d x$
Solution:
Put

\begin{aligned} e^{\sqrt{x}} = t \Rightarrow x = (\ln t)^{2} \Rightarrow d x = 2 \frac{\ln t}{t} d t \\ ∴ \int e^{\sqrt{x}} d x \Rightarrow 2 \int t \frac{\ln t}{t} d t = 2 \int \ln t d t \end{aligned}

[Now integrate by parts]

\Rightarrow 2 [t \ln t - \int t d (\ln t)] \Rightarrow 2 [t - \int t \frac{d t}{t}]

\begin{aligned} = 2 [t \ln t - \int d t] = 2 [t \ln t - t + c] \\ = 2 [t (\ln t - 1) + c] \\ = 2 e^{\sqrt{x}} (\sqrt{x} - 1) + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 23

Answer:
$= \frac{- [\log (x + 2) + 1]}{x + 2} + c$
Hint: Let
$x + 2 = t$
Given: $\int \frac{\log (x + 2)}{(x + 2)^{2}} d x$
Solution:
Let

x + 2 = t \Rightarrow d x = d t

\begin{aligned} \int \frac{\log (x + 2)}{(x + 2)^{2}} d x \Rightarrow \int \frac{\log t}{t^{2}} d t \Rightarrow \int \log t \frac{1}{t^{2}} d t \\ = \log t \int \frac{1}{t^{2}} d t - \int \frac{d}{d t} \log t (\int \frac{1}{t^{2}} d t) d t + c \\ = \log t (\frac{- 1}{t}) - \int \frac{1}{t} (\frac{- 1}{t}) d t + c \\ = - \frac{\log t}{t} + \int \frac{1}{t^{2}} d t + c \\ = - \frac{\log t}{t} + (\frac{- 1}{t}) + c \\ = \frac{- (\log t + 1)}{t} + c \\ = \frac{- [\log (x + 2) + 1]}{x + 2} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 24

Answer: $x \tan \frac{x}{2} + c$
Hint: Use trigonometric formulae
Given: $\int \frac{x + \sin x}{1 + \cos x} d x$
Solution:
$\begin{aligned} \int (\frac{x}{1 + \cos x} + \frac{\sin x}{1 + \cos x}) d x \\ \Rightarrow \int \frac{x}{2 \cos^{2} \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}] d x \\ \Rightarrow \frac{1}{2} \int x \sec^{2} \frac{x}{2} + \int \tan \frac{x}{2} d x \\ \Rightarrow \frac{1}{2} [x \frac{\tan \frac{x}{2}}{\frac{1}{2}} - \int 1 \tan \frac{x}{2} \times 2] + \int \tan \frac{x}{2} d x \end{aligned}$
$\begin{aligned} \Rightarrow x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x + \int \tan \frac{x}{2} d x \\ \Rightarrow x \tan \frac{x}{2} + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 25

Answer: $(\frac{x}{\log 10}) [\log x - 1] + c$
Hint: Take
$\log x = \frac{\log x}{\log 10}$
Given: $\int \log_{10} x d x$
Solution:
$\begin{aligned} \int \log_{10} x d x = \int (\frac{\log x}{\log 10}) d x \\ = (\frac{1}{\log 10}) \int \log x \cdot 1 d x \\ = (\frac{1}{\log 10}) [(\log x) x - \int \frac{1}{x} x d x] \\ = (\frac{1}{\log 10}) (x \log x - x) + c \\ = x [(\frac{\log x}{\log 10}) - (\frac{1}{\log 10})] + c \\ = (\frac{x}{\log 10}) [\log x - 1] + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 26

Answer: $2 [\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}]$
Hint: $\sqrt{x} = t$
Given: $\int \cos \sqrt{x d x}$
Solution: Put $\sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t$
$∴ \int \cos \sqrt{x} d x = 2 \int t \cos t d t$
[Using integration by parts]

\begin{aligned} I = 2 [t \sin t - \int 1 \sin t d t] \\ = 2 [t \sin t + \cos t] \\ = 2 [\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 27

Answer: $- [\sqrt{1 - x^{2}} \cos^{- 1} x + x] + c$
Given: $\int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x$
Hint: Use integration by parts
Solution:
$I = \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x \Rightarrow - \frac{1}{2} \int \frac{- 2 x}{\sqrt{1 - x^{2}}} \cos^{- 1} x d x$
[Using integration by parts]

\begin{aligned} = \frac{- 1}{2} [\cos^{- 1} x \times \int \frac{- 2 x}{\sqrt{1 - x^{2}}} d x - \int {(\frac{d}{d x} \cos^{- 1} x) \int \frac{- 2 x}{\sqrt{1 - x^{2}}} d x} d x] \\ = \frac{- 1}{2} [\cos^{- 1} x \cdot 2 \sqrt{1 - x^{2}} - \int - \frac{1}{\sqrt{1 - x^{2}}} 2 \sqrt{1 - x^{2}} d x] \\ = \frac{- 1}{2} [2 \sqrt{1 - x^{2}} \cos^{- 1} x + 2 x] + c \\ = - [\sqrt{1 - x^{2}} \cos^{- 1} x + x] + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 28

Answer: $- \frac{\log x}{(x + 1)} + \log x - \log (x + 1)$
Hint:
Given: $\int \frac{l o g x}{{(x + 1)}^{2}} d x$
Solution:
$I = \frac{l o g x}{{(x + 1)}^{2}} d x$
$= l o g x \int \frac{1}{{(x + 1)}^{2}} d x - \int (\frac{1}{x} \int \frac{1}{{(x + 1)}^{2}} d x) d x$
$= l o g x (\frac{- 1}{x + 1}) + \int \frac{1}{x} \frac{1}{x + 1} d x$
$= - \frac{l o g x}{(x + 1)} + I_{1}$
$= I_{1} = \int \frac{1}{x (x + 1)} d x$
Using partial fractions
Let

\frac{1}{x (x + 1)} = \frac{A}{x} + \frac{B}{x + 1}

1 = A (x + 1) + B (x)

1 = A x + A + B x

1 = (A + B) x + A

\Rightarrow A + B = 0

\Rightarrow A = 1

\Rightarrow B = - 1

I_{1} = \int [\frac{1}{x} - \frac{1}{x + 1}] d x

= l o g x - l o g (x + 1)

I = \frac{l o g x}{x + 1} + l o g x - l o g (x + 1)

Indefinite Integrals Exercise 18.25 Question 29

Answer: $\frac{- 1}{2} \cos e c x \cot x + \frac{1}{2} l o g | \tan \frac{x}{2} | + c$
Hint: $\cos e c^{3} x = \cos e c^{2} x \cos e c x$
Given: $\int \cos e c^{3} x d x$
Solution:
Let
$\begin{aligned} I = \int {cosec}^{3} d x \\ = \int {cosec}^{2} x \cdot cosec x d x \\ = \int {cosec}^{2} x \cdot \sqrt{1 + \cot^{2} x} \end{aligned}$
$l e t \cot x = t$
$\begin{aligned} I = - \int \sqrt{1 + t^{2}} d t \\ = - \frac{t}{2} \sqrt{1 + t^{2}} - \frac{1^{2}}{2} \log | t + \sqrt{1 + t^{2}} | + C \end{aligned}$
Substituting the value of t in equation (1)

\begin{aligned} = - \frac{\cot x}{2} \cdot \cos e c x - \frac{1}{2} \log | \cot x + \cos e c x | + C \\ = - \frac{1}{2} \cos e \cot x - \frac{1}{2} \log | \frac{\cos x}{\sin x} + \frac{1}{\sin x} | + C \\ = - \frac{1}{2} \cos e c x \cot x - \frac{1}{2} \log | \frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}} | + C \\ = - \frac{1}{2} \cos e cx \cot x - \frac{1}{2} \log | \cot \frac{x}{2} | + C \\ = - \frac{1}{2} \cos e c x \cot x + \frac{1}{2} \log | \tan \frac{x}{2} | + C [∵ \log | \cot \frac{x}{2} | = \log | \frac{1}{\tan \frac{x}{2}} | \Rightarrow - \log | \tan \frac{x}{2} |] \end{aligned}

Indefinite Integrals Exercise 18.25 Question 30

Answer: $\frac{- \cos^{4 (\cos^{- 1} \sqrt{x})}}{2} + c$
Hint: Let
$x = \cos^{2} t$
Given:
$\int \sec^{- 1} \sqrt{x d x}$ ................(1)
Solution:
Assume that $\sqrt{x} = t$ ...............(ii)
$\Rightarrow x = t^{2}$
Differentiating w.r.t to ‘t’

d x = 2 t d t

.................(iii)
Now substituting values from equation (ii) & equation (iii)

$\begin{aligned} I & = \int \sec^{- 1} t (2 t d t) \\ I & = 2 \int \sec^{- 1} t . t d t \\ = 2 {\sec^{- 1} t \int t d t - \int (\frac{d (\sec^{- 1} t)}{d t} (\int t d t)) d x} \end{aligned}$
Here, use integration Identity
$[\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$

$\begin{matrix} I = 2 {\sec^{- 1} t \frac{t^{2}}{2} - \int (\frac{1}{t \sqrt{t^{2} - 1}} (\frac{t^{2}}{2})) d t} + c \\ I = t^{2} \sec^{- 1} t - \int \frac{t}{\sqrt{t^{2} - 1}} d t + c \end{matrix}$ ...........(iv)
Assume

$t^{2} - 1 = y$ ...........(v)
Differentiate w.r.t ‘y’

$2 t d t = d y$

$t d t = \frac{d y}{2}$ ............(vi)
Substituting values from equation (v) and equation (vi) in integral of equation (iv)

$\begin{aligned} I = t^{2} \sec^{- 1} t - \int \frac{t}{\sqrt{y}} \frac{d y}{2} + c \\ I = t^{2} \sec^{- 1} t - \frac{1}{2} \int y^{- \frac{1}{2}} d y + c \end{aligned}$
Use integration identity

$[\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]$

$\begin{aligned} I = t^{2} \sec^{- 1} t - \frac{1}{2} \frac{y^{\frac{1}{2}}}{\frac{1}{2}} + c \\ I = t^{2} \sec^{- 1} t - \sqrt{y} + c \end{aligned}$
Substitute the value of y from equation (v)

$I = t^{2} \sec^{- 1} t - \sqrt{t^{2} - 1} + c$
Substitute the value of ‘t’ from equation (ii)

$I = x \sec^{- 1} \sqrt{x} - \sqrt{x - 1} + c$

So,

$\int \sec^{- 1} \sqrt{x} d x = x \sec^{- 1} \sqrt{x} - \sqrt{x - 1} + c$

Indefinite Integrals Exercise 18.25 Question 31

Answer: $x \sin^{- 1} (\sqrt{x}) + \frac{1}{2} [\sqrt{1 - x} \sqrt{x} + \sin^{- 1} \sqrt{1 - x}] + c$
Given: $\int \sin^{- 1} \sqrt{x d x}$
Hint: $u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Solution:
$\begin{aligned} u = \int \sin^{- 1} \sqrt{x} d x \\ = \int \sin^{- 1} \sqrt{x} \cdot 1 d x \\ = \sin^{- 1} \sqrt{x} \int 1 d x - \int (\frac{d}{d x} (\sin^{- 1} \sqrt{x}) \int 1 d x) d x \end{aligned}$
$\begin{aligned} [\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}] \\ = x \sin^{- 1} (\sqrt{x}) - \int \frac{1}{\sqrt{1 - (\sqrt{x})^{2}}} \times \frac{1}{2 \sqrt{x}} x d x \\ = x \sin^{- 1} (\sqrt{x}) - \int \frac{\sqrt{x}}{2 \sqrt{1 - (\sqrt{x})^{2}}} d x \end{aligned}$
Now,
$\sqrt{1 - {(\sqrt{x})}^{2}} = t$
$\Rightarrow \sqrt{1 - x} = t$
$\Rightarrow 1 - x = t^{2}$
$\Rightarrow 1 - t^{2} = x$
$\Rightarrow \sqrt{x} = \sqrt{1 - t^{2}}$
Again
$\Rightarrow 1 - x = t^{2}$
$- d x = 2 t d t$
$d x = - 2 t d t$
$[\int \sqrt{a^{2} - x^{2}} d x = \frac{1}{2} [x \sqrt{a^{2} - x^{2}} a^{2} \sin^{- 1} \frac{x}{a}]]$
$\begin{aligned} = x \sin^{- 1} (\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{1 - t^{2}}}{t} (- 2 t d t) \\ = x \sin^{- 1} (\sqrt{x}) + \frac{1}{2} \times 2 \int \sqrt{1 - t^{2}} d t \\ = x \sin^{- 1} (\sqrt{x}) + \frac{1}{2} [t \sqrt{1 - t^{2}} + 1 \sin^{- 1} t] + c \\ = x \sin^{- 1} (\sqrt{x}) + \frac{1}{2} [\sqrt{1 - x} \sqrt{x} + \sin^{- 1} \sqrt{1 - x}] + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 32

Answer: $x \tan x + l o g (\cos x) - \frac{x^{2}}{2} + c$
Given: $\int x \tan^{2} x d x$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Solution:
$\begin{aligned} I = \int x \tan^{2} x d x \\ [\tan^{2} x = \sec^{2} x - 1] \\ I = \int x \sec^{2} x d x - \int x d x \\ I = x \int \sec^{2} x d x - \int [\frac{d}{d x} x \int \sec^{2} x d x] d x - \frac{x^{2}}{2} + c \\ I = x \tan x - \int \tan x d x - \frac{x^{2}}{2} + c \\ I = x \tan x + \log (\cos x) - \frac{x^{2}}{2} + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 33

Answer: $= \tan x + l o g (\cos x) - \frac{x^{2}}{2} + c$
Given: $\int x (\frac{\sec 2 x - 1}{\sec 2 x + 1}) d x$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Solution:
$\begin{aligned} I = \int x (\frac{1 - \cos 2 x}{1 + \cos 2 x}) d x \\ I = \int x (\frac{2 \sin^{2} x}{2 \cos^{2} x}) d x \\ I = \int x (\sec^{2} x - 1) d x \\ I = \int x \sec^{2} x - \int x d x \\ I = x \int \sec^{2} x - \int [\frac{d}{d x} x \int \sec^{2} x d x] d x - \frac{x^{2}}{2} + c \\ I = x \tan x - \int \tan x d x - \frac{x^{2}}{2} + c \\ = x \tan x + \log (\cos x) - \frac{x^{2}}{2} + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 34

Answer: $x e^{x} [l o g (x e^{x}) - 1] + c$
Hint: $t = x . e^{x}$
Given: $\int (x + 1) e^{x} l o g (x . e^{x}) d x$
Solution:
Now,Let

t = x . e^{x}

d t = (x . e^{x} + e^{x} 1) d x

d t = (x + 1) e^{x} d x

\begin{aligned} ∴ I = \int \log t d t \\ I = \int \log (t) \cdot 1 d t \\ I = \log t \int 1 d t - \int \frac{1}{t} t d t \\ I = t \log t - \int d t \end{aligned}

I = t l o g t - t + c

= t (l o g t - 1) + c

= x e^{x} [l o g (x e^{x}) - 1] + c

Indefinite Integrals Exercise 18.25 Question 35

Answer: $3 \sin^{- 1} x, x + 3 \sqrt{1 - x^{2}} + c$
Given: $\int \sin^{- 1} (3 x - 4 x^{3}) d x$
Hint:
$3 x - 4 x^{3} \to \sin θ$
$x = \sin t$
Solution:
$I = \int \sin^{- 1} (3 x - 4 x^{3}) d x$
Let x = sint =>dx = costdt &

3 \sin t - 4 \sin^{3} t = \sin 3 t

\begin{aligned} = \int \sin^{- 1} \sin 3 t \cos t d t \\ = \int 3 t \cos t d t \\ = 3 \int t \cos t d t \\ I = 3 [(t \int \cos t d t - \int (1 \sin t) d t)] \\ = 3 [t \sin t - \int \sin t d t] \\ = 3 [t \sin t - (- \cos t + c)] \\ = 3 t \sin t + 3 \cos t + c \\ = 3 \sin^{- 1} x \cdot x + 3 \sqrt{1 - x^{2}} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 36

Answer: $2 x \tan^{- 1} (x) - l o g (1 + x^{2}) + c$
Given: $\int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x$
Hint: $2 \tan^{- 1} x = \sin^{- 1} x \frac{2 x}{1 + x^{2}}$
Solution:
$\begin{aligned} I = \int 2 \tan^{- 1} x d x \\ I = 2 x \tan^{- 1} (x) - \int \frac{1}{1 + x^{2}} 2 x d x \\ 1 + x^{2} = t \Rightarrow 2 x d x = d t \\ I = 2 x \tan^{- 1} (x) - \int \frac{1}{t} d t \\ I = 2 x \tan^{- 1} (x) - \log (t) + c \\ I = 2 x \tan^{- 1} (x) - \log (1 + x^{2}) + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 38

Answer: $\frac{x^{3}}{3} \sin^{- 1} x + \frac{1}{3} \sqrt{1 - x^{2}} - \frac{1}{9} {(1 - x^{2})}^{\frac{3}{2}} + c$
Given: $\int x^{2} \sin^{- 1} x d x$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Solution:
$I = \int x^{2} \sin^{- 1} x d x$
$\begin{aligned} I = (\sin^{- 1} x) \frac{x^{3}}{3} - \int \frac{1}{3 \sqrt{1 - x^{2}}} \times x^{3} d x \dots applying byparts \\ I = \frac{x^{3}}{3} \sin^{- 1} x - \frac{1}{3} \int \frac{x^{2} x}{\sqrt{1 - x^{2}}} d x \end{aligned}$
Let 1-

x^{2}

=t => -2xdx = dt

\begin{aligned} = \frac{x^{3}}{3} \sin^{- 1} x + \frac{1}{2} \times \frac{1}{3} \int \frac{(1 - t) d t}{\sqrt{t}} \\ = \frac{x^{3}}{3} \sin^{- 1} x + \frac{1}{6} [\int \frac{1}{\sqrt{t}} d t - \int \frac{t}{\sqrt{t}} d t] \\ = \frac{x^{3}}{3} \sin^{- 1} x + \frac{1}{6} [2 \sqrt{t} - \frac{2 t^{\frac{3}{2}}}{3}] \\ = \frac{x^{3}}{3} \sin^{- 1} x + \frac{1}{3} \sqrt{1 - x^{2}} - \frac{1}{9} {(1 - x^{2})}^{\frac{3}{2}} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 39

Answer: $\frac{- \sin^{- 1} x}{x} + \frac{1}{2} [l o g \frac{\sqrt{1 - x^{2} - 1}}{1 + x^{2} + 1}] + c$
Given: $\int \frac{s i n^{- 1} x}{x^{2}} d x$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Solution:
$\begin{aligned} I = \int \frac{\sin^{- 1} x}{x^{2}} d x \\ \int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x \\ I = \sin^{- 1} \int \frac{1}{x^{2}} d x - [d \frac{\sin^{- 1} x}{d x} (\int \frac{1}{x^{2}} d x) d x] \\ I = \frac{- \sin^{- 1} x}{x} + \int \frac{d x}{x \sqrt{1 - x^{2}}} \end{aligned}$
Let

\sqrt{1 - x^{2}} = u

\Rightarrow 1 - x^{2} = u^{2}

\Rightarrow \frac{- 2 x}{2 \sqrt{1 - x^{2}}} d x = d u

\Rightarrow \frac{- x d x}{\sqrt{1 - x^{2}}} = d u

∴ - d u = \frac{x d x}{\sqrt{1 - x^{2}}}

Now,

\begin{aligned} y = \int \frac{d x}{x \sqrt{1 - x^{2}}} \\ \Rightarrow \int \frac{x d x}{x^{2} \sqrt{1 - x^{2}}} \\ \Rightarrow - \int \frac{d u}{1 - u^{2}} \\ \Rightarrow \frac{d u}{u^{2} - 1} \\ \Rightarrow \int \frac{d u}{(u + 1) (u - 1)} \Rightarrow \frac{1}{2} \int \frac{2 d u}{(u + 1) (u - 1)} \end{aligned}

\begin{aligned} \Rightarrow \frac{1}{2} \int \frac{(u + 1) - (u - 1)}{(u + 1) (u - 1)} d u \\ \Rightarrow \frac{1}{2} [\int \frac{(u + 1)}{(u + 1) (u - 1)} d u - \int \frac{(u - 1)}{(u + 1) (u - 1)} d u] \\ \Rightarrow \frac{1}{2} [\log (u - 1) - \log (u + 1)] \\ \Rightarrow \frac{1}{2} [\log \frac{\sqrt{1 - x^{2}} - 1}{\sqrt{1 - x^{2}} + 1}] + c \\ I = \frac{- \sin^{- 1} x}{x} + \int \frac{d x}{x \sqrt{1 - x^{2}}} \\ = \frac{- \sin^{- 1} x}{x} + \frac{1}{2} [\log \frac{\sqrt{1 - x^{2}} - 1}{\sqrt{1 - x^{2}} + 1}] + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 40

Answer: $x \tan^{- 1} x - \frac{1}{2} \ln (1 + x^{2}) + \frac{{(t a n^{- 1} x)}^{2}}{2} + c$
Given: $\int \frac{x^{2} \tan^{- 1} x}{1 + x^{2}} d x$
Hint: Let

1 + x^{2} = u

and

\tan^{- 1} x = t

Solution:
$\begin{aligned} \int \frac{x^{2} \tan^{- 1} x}{1 + x^{2}} d x \\ = \int \frac{(x^{2} + 1 - 1) \tan^{- 1} x}{1 + x^{2}} d x \\ = \int (\frac{1 + x^{2}}{1 + x^{2}}) \tan^{- 1} x - \int \frac{\tan^{- 1} x}{1 + x^{2}} d x \\ = \int \tan^{- 1} x d x - \int \frac{\tan^{- 1} x}{1 + x^{2}} d x \\ \int 1 \tan^{- 1} x = x \tan^{- 1} x - \int \frac{x}{1 + x^{2}} d x \end{aligned}$
Let

1 + x^{2} = u \Rightarrow 2 x d x = d x

\begin{aligned} \int 1 \tan^{- 1} x = x \tan^{- 1} x - \int \frac{d u}{2 u} \\ \Rightarrow x \tan^{- 1} x - \frac{1}{2} \ln u \\ \Rightarrow x \tan^{- 1} x - \frac{1}{2} \ln (1 + x^{2}) \end{aligned}

\int \frac{\tan^{- 1} x}{1 + x^{2}} d x

Let,

\tan^{- 1} x =\Rightarrow d t = \frac{1}{1 + x^{2}} d x

\begin{aligned} \int t d t = \frac{t^{2}}{2} = \frac{{(\tan^{- 1} x)}^{2}}{2} \\ \Rightarrow \int x d x - \int \frac{x}{1 + x^{2}} d x = x \tan^{- 1} x - \frac{1}{2} \ln (1 + x^{2}) - \frac{{(\tan^{- 1} x)}^{2}}{2} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 41

Answer: $3 x \cos^{- 1} x - 3 \sqrt{1 - x^{2}} + c$
Hint: $\cos 3 x = 4 \cos^{3} x - 3 \cos x$
Given: $\int \cos^{- 1} (4 x^{3} - 3 x) d x$
Solution: $I = \int \cos^{- 1} (4 x^{3} - 3 x) d x$
Let

x = \cos θ \Rightarrow θ = \cos^{- 1} x

\begin{aligned} I = \int \cos^{- 1} (4 \cos^{3} θ - 3 \cos θ) d x \\ = \int \cos^{- 1} (\cos 3 θ) d x \\ = \int 3 θ d x \\ = 3 \int \cos^{- 1} x d x \end{aligned}

\begin{aligned} = 3 [\cos^{- 1} x (x) - \frac{1}{2} \int u^{\frac{- 1}{2}} d u] \\ = 3 [x \cos^{- 1} x - \frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}}] + c \\ = 3 x \cos^{- 1} x - 3 \sqrt{1 - x^{2}} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 42

Answer: $2 x \tan^{- 1} x - 2 l o g (\sqrt{1 + x^{2}}) + c$
Given: $\int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x$
Hint: Put $x = \tan Θ$
Solution:
$I = \int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x$
Put

x = \tan θ \Rightarrow d x = \sec^{2} θ d θ

\begin{aligned} I = \int \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \sec^{2} θ d θ \\ [\cos 2 θ = (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ})] \\ I = \int \cos^{- 1} (\cos 2 θ) \sec^{2} θ d θ \end{aligned}

\begin{aligned} = \int 2 θ \sec^{2} θ d θ \\ = 2 \int θ \sec^{2} θ d θ \\ = 2 [θ \int \sec^{2} θ d θ - \int \frac{d}{d x} θ (\int \sec^{2} θ d θ) d θ] \\ = 2 θ \times \tan θ - 2 \int \tan θ d θ \\ = 2 θ \tan θ - 2 θ + c \\ θ = \tan^{- 1} x \\ = 2 x \tan^{- 1} x - 2 \log (\sqrt{1 + x^{2}}) + c \\ = 2 x \tan^{- 1} x - 2 \log (\sqrt{1 + x^{2}}) + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 43

Answer: $2 x \tan^{- 1} x - l o g (1 + x^{2}) + c$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Given: $\int \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x$
Solution:
$\begin{aligned} I = 2 \int \tan^{- 1} x d x \\ = 2 [\tan^{- 1} x \int d x - \int [\frac{d (\tan^{- 1} x)}{d x} \int d x] d x] \\ = 2 [x \tan^{- 1} x - \int \frac{x}{1 + x^{2}} d x] \end{aligned}$
$\int \frac{x}{1 + x^{2}} d x$
Put

1 + x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2}

\begin{aligned} \Rightarrow \int \frac{x}{1 + x^{2}} d x = \frac{1}{2} \int \frac{d t}{t} \\ = \frac{1}{2} \log t + c \\ = \frac{1}{2} \log (1 + x^{2}) + c \end{aligned}

2 [x \tan^{- 1} x - \int \frac{x}{1 + x^{2}} d x] = 2 x \tan^{- 1} x - \log (1 + x^{2}) + c

Indefinite Integrals Exercise 18.25 Question 44

Answer: $l o g x (\frac{x^{2}}{2} + x) - \frac{x^{2}}{4} - x + c$

Given: $\int (x + 1) l o g x d x$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Solution:
$\begin{aligned} I = \int (x + 1) \log x d x \\ I = [\log x \int (x + 1) d x - \int \frac{1}{x} (\frac{x^{2}}{2} + x) d x] \\ I = \log x (\frac{x^{2}}{2} + x) - \int (\frac{x}{2} + 1) d x \\ = \log x (\frac{x^{2}}{2} + x) - \int \frac{x}{2} d x - \int d x \\ = \log x (\frac{x^{2}}{2} + x) - \frac{x^{2}}{4} - x + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 45

Answer: $\frac{x^{3}}{3} \tan^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} l o g | 1 + x^{2} | + c$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Given: $\int x^{2} \tan^{- 1} x d x$
Solution:
$\begin{aligned} \int x^{2} \tan^{- 1} x d x \\ = \tan^{- 1} x \int x^{2} d x - \int \frac{d}{d x} \tan^{- 1} x (\int x^{2} d x) \\ = \frac{x^{3}}{3} \tan^{- 1} x - \frac{1}{3} \int \frac{x (1 + x^{2})}{1 + x^{2}} d x + \frac{1}{6} \int \frac{2 x}{1 + x^{2}} d x \\ = \frac{x^{3}}{3} \tan^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} \log | 1 + x^{2} | + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 46

Answer: $x \sin x + \cos x - \frac{1}{4} \cos 2 x + c$
Given: $\int (e^{l o g x} + \sin x) \cos x d x$
Hint: Put

e^{l o g x} = x

Solution: $\int (e^{l o g x} + \sin x) \cos x d x$
As

e^{l o g x} = x

\begin{aligned} I = \int (x + \sin x) \cos x d x \\ = \int x \cos x d x + \int \sin x \cos x d x \\ = x \sin x - \int 1 \sin x d x + \frac{1}{2} \int \sin (2 x) d x \\ = x \sin x - (- \cos x) + \frac{1}{2} \frac{1}{2} (- \cos 2 x) + c \\ = x \sin x + \cos x - \frac{1}{4} \cos 2 x + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 47

Answer: $\frac{1}{\sqrt{1 + x^{2}}} (x - \tan^{- 1} x) + c$
Hint: $x = \tan θ$ & $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Given: $\int \frac{x \tan^{- 1} x}{{(1 + x^{2})}^{\frac{3}{2}}} d x$
Solution:
$I = \frac{x \tan^{- 1} x}{{(1 + x^{2})}^{\frac{3}{2}}} d x$
Put

x = \tan Θ \Rightarrow d x = \sec^{2} Θ d Θ

\begin{matrix} I = \int \frac{\tan θ \tan^{- 1} (\tan θ) \sec^{2} θ d θ}{{(\sqrt{1 + \tan^{2} θ})}^{\frac{3}{2}}} \\ = \int \frac{\tan \sec^{2} θ}{\sqrt{\sec^{2} θ} \sec^{3} θ} \\ = \int (\frac{θ \tan θ}{\sec θ}) d θ \end{matrix}

\begin{aligned} I = \int θ \frac{\sin θ}{\cos θ} \times \cos θ d θ \\ I = θ \int \sin θ d θ - \int [\frac{d}{d θ} θ \int \sin θ d θ] d θ \\ = - θ \cos θ + \int (\cos θ) d θ \\ = - θ \cos θ + \sin θ + c \\ = - \tan^{- 1} x (\frac{1}{\sqrt{1 + x^{2}}}) + \frac{x}{\sqrt{1 + x^{2}}} + c \\ = \frac{1}{\sqrt{1 + x^{2}}} (x - \tan^{- 1} x) + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 48

Answer: $\tan^{- 1} \sqrt{x} (1 + x) - \sqrt{x} + c$
Given: $\int (\tan^{- 1} \sqrt{x}) d x$
Hint:Put

\sqrt{x} = t

Solution:
$I = (\tan^{- 1} \sqrt{x}) d x$
Put

\sqrt{x} = t \Rightarrow d x = 2 t d t

\begin{aligned} I = \int \tan^{- 1} t 2 t d t \\ I = \tan^{- 1} t 2 \cdot \frac{t^{2}}{2} - \int \frac{t^{2} + 1 - 1}{1 + t^{2}} d t \\ = t^{2} \tan^{- 1} t - \int d t + \int \frac{1}{1 + t^{2}} d t \\ = t^{2} \tan^{- 1} t - t + \tan^{- 1} t + c \\ = \tan^{- 1} t (1 + t^{2}) - t + c \\ = \tan^{- 1} \sqrt{x} (1 + x) - \sqrt{x} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 49

Answer: $\frac{x^{3}}{3} \tan^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} l o g (1 + x^{2}) + c$
Hint: $\int u v d x = u \int v d x - \int (\frac{d}{d x} u \int v d x) d x$
Given: $\int x^{2} \tan^{- 1} x d x$
Solution:
$\int x^{2} \tan^{- 1} x d x$
$\begin{aligned} = \tan^{- 1} x \int x^{2} d x - \int \frac{d}{d x} \tan^{- 1} x (\int x^{2} d x) d x \\ = \frac{x^{3}}{3} \tan^{- 1} x - \frac{1}{3} \int \frac{x \cdot x^{2}}{1 + x^{2}} d x \\ = \frac{x^{3}}{3} \tan^{- 1} x - \frac{1}{3} \int \frac{x (1 + x^{2})}{1 + x^{2}} d x + \frac{1}{6} \int \frac{2 x}{1 + x^{2}} d x \\ = \frac{x^{3}}{3} \tan^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} \log (1 + x^{2}) + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 50

Answer: $\frac{\sin 3 x}{18} - \frac{x \cos 3 x}{6} + \frac{x \cos x}{2} - \frac{\sin x}{2} + c$
Hint: $\int s i n x d x = \cos x$ & $\int \cos x d x = - \sin x$
Given: $\int x \sin x \cos 2 x d x$
Solution:
$I = \int x \sin x \cos 2 x d x$
$I = \frac{1}{2} \int 2 x \sin x \cos 2 x d x$
$= \frac{1}{2} \int x 2 \sin x \cos 2 x d x$
$\begin{aligned} = \frac{1}{2} \int x [\sin (x + 2 x) + \sin (x - 2 x)] d x \\ = \frac{1}{2} \int x \sin 3 x d x - \frac{1}{2} \int x \sin x d x \\ = \frac{1}{2} [x \frac{\cos 3 x}{3} + \frac{1}{3} \int \cos 3 x d x] - \frac{1}{2} [x \cos x + \int \cos x d x] \\ I = \frac{1}{2} [- x \frac{\cos 3 x}{3} + \frac{\sin 3 x}{9}] - \frac{1}{2} [- x \cos x + \sin x] \\ = \frac{1}{2} [\frac{\sin 3 x}{9} - \frac{x \cos 3 x}{3} + x \cos x - \sin x] + c \\ = \frac{\sin 3 x}{18} - \frac{x \cos 3 x}{6} + \frac{x \cos x}{2} - \frac{\sin x}{2} + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 51

Answer: $\frac{1}{2} x^{2} \tan^{- 1} x^{2} - \frac{1}{4} l o g (1 + x^{4}) + c$
Hint: Put $x^{2} = t$
Given: $\int x (\tan^{- 1} x^{2}) d x$
Solution:By letting

x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2}

∴ \int x \tan^{- 1} (x^{2}) d x = \frac{1}{2} \int \tan^{- 1} t .1 d t

Integrate by parts, taking

\tan^{- 1} t

as the first function

\begin{aligned} = \frac{1}{2} [\tan^{- 1} t . t - \int \frac{1}{1 + t^{2}} t d t] \\ = \frac{1}{2} [t \tan^{- 1} t - \frac{1}{2} \int \frac{2 t}{1 + t^{2}} d t] \\ = \frac{1}{2} [t \tan^{- 1} t - \frac{1}{2} \log | 1 + t^{2} |] + c \\ = \frac{1}{2} x^{2} \tan^{- 1} x^{2} - \frac{1}{4} \log (1 + x^{4}) + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 52

Answer: $- \sqrt{1 - x^{2}} \sin^{- 1} x + x + c$
Hint: Put $\sin^{- 1} x = t$
Given: $\int \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x$
Solution:By letting

\sin^{- 1} x = t \Rightarrow x = \sin t \Rightarrow d x = \cos t d t

\begin{aligned} ∴ \int \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x & = \int \frac{\sin t . t}{\sqrt{1 - \sin^{2} t}} \cos t d t \\ = \int \frac{t \sin t \cos t}{\cos t} d t \\ = \int t \sin t d t \Rightarrow t (- \cos t) - \int 1 (- \cos t) d t \end{aligned}

Applying by parts

= t \cos t + \sin t + c

= t \sqrt{1 - t} + \sin t + c

= \sqrt{1 - x^{2}} \sin^{- 1} x + x + c

Indefinite Integrals Exercise 18.25 Question 54

Answer: $- \frac{3}{2} \sqrt{x} \cos \sqrt{x} + \frac{3}{2} \sin \sqrt{x} + \frac{1}{6} [\sqrt{x} \cos 3 \sqrt{x} - \frac{\sin 3 \sqrt{x}}{3}] + c$
Hint: Put

\sqrt{x} = t

Given: $\int \sin^{3} \sqrt{x d x}$
Solution: Put

\sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x}

\Rightarrow d x = 2 t d t

\begin{aligned} ∴ \int \sin^{3} \sqrt{x} d x = 2 \int t \sin^{3} t d t \\ [\begin{matrix} ∴ \sin 3 t = 3 \sin t - 4 \sin^{3} t \\ \sin^{3} t = \frac{3 \sin t - \sin 3 t}{4} \end{matrix}] \\ \Rightarrow 2 \int \frac{t [3 \sin t - \sin 3 t]}{4} d t \\ \Rightarrow \frac{3}{2} \int t \sin t d t - \frac{1}{2} \int t \sin 3 t d t \end{aligned}

Applying by parts,

$\begin{aligned} = \frac{3}{2} [t (- \cos t) - \int (- \cos t) d t] - \frac{1}{2} [t (\frac{- \cos 3 t}{3}) - \int (\frac{- \cos 3 t}{3}) d t] \\ = \frac{3}{2} [- t \cos t + \sin t] - \frac{1}{2} [\frac{- t \cos 3 t}{3} + \frac{1}{3} \frac{\sin 3 t}{3}] + c \\ = - \frac{3}{2} \sqrt{x} \cos \sqrt{x} + \frac{3}{2} \sin \sqrt{x} + \frac{1}{6} [\sqrt{x} \cos 3 \sqrt{x} - \frac{\sin 3 \sqrt{x}}{3}] + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 55

Answer: $\frac{- 3 x \cos x}{4} + \frac{3 \sin x}{4} + \frac{x \cos 3 x}{12} - \frac{\sin 3 x}{36} + c$
Hint:
Given: $\int x \sin^{3} x d x$
Solution: $∴ \sin 3 x = 3 \sin x - 4 \sin^{3} x$
$\sin^{3} x = \frac{3 \sin x - \sin 3 x}{4} d x$
$= \frac{1}{4} \int 3 x \sin x - x \sin 3 x d x$
$= \frac{3}{4} \int x \sin x - \frac{1}{4} \int x \sin 3 x d x$
Take the first function as x and second function as sinx and sin 3x

\begin{aligned} = \frac{3}{4} [x \int \sin x - \int [\frac{d x}{d x} \int \sin x d x] d x] - \frac{1}{4} [x \int (\sin 3 x d x) - \int [\frac{d x}{d x} \int \sin 3 x d x] d x] \\ = \frac{3}{4} (- x \cos x + \int \cos x d x) - \frac{1}{4} [\frac{- x \cos 3 x}{3} + \int \frac{\cos 3 x}{3} d x] \\ = \frac{3}{4} (- x \cos x + \sin x) - \frac{1}{4} [\frac{- x \cos 3 x}{3} + \frac{\sin 3 x}{9}] + c \\ = \frac{- 3 x \cos x}{4} + \frac{3 \sin x}{4} + \frac{x \cos 3 x}{12} - \frac{\sin 3 x}{36} + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 56

Answer: $= 2 \sqrt{x} \sin \sqrt{x} + \frac{4}{3} \cos \sqrt{x} - \frac{2 \sqrt{x} \sin^{3} \sqrt{x}}{3} + \frac{2}{9} [\cos^{3} \sqrt{x}] + c$
Hint: Let

\sqrt{x} = t

Given: $\int \cos^{3} \sqrt{x d x}$
Solution: $\sqrt{x} = t \Rightarrow x = t^{2} \Rightarrow d x = 2 t d t$
$\begin{aligned} ∴ \int \cos^{3} \sqrt{x} d x = \int \cos^{3} t .2 t d t \Rightarrow 2 \int t \cos^{3} t d t \\ = 2 \int t \cos t \cos^{2} t d t \Rightarrow 2 \int t \cdot \cos t (1 - \sin^{2} t) d t \\ = 2 \int (t \cos t - t \cos t \sin^{2} t) d t \\ = 2 [\int t \cos t d t - \int t \cos t \sin^{2} t d t] \\ = 2 [t \int \cos t d t - \int [1 \cdot \sin t d t]] - 2 [t \int \cos \sin^{2} t d t - \int 1 \cdot (\frac{\sin^{3} t}{3}) d t] \end{aligned}$
Applying by parts

\begin{aligned} = 2 (t \sin t - \int \sin t d t) - 2 (t \cdot \frac{\sin^{3} t}{3} - \frac{1}{3} \int \sin^{3} t d t) \\ = 2 t \sin t + 2 \cos t - \frac{2 t \sin^{3} t}{3} + \frac{2}{3} \int \sin t \cdot \sin^{2} t d t \\ = 2 t \sin t + 2 \cos t - \frac{2 t \sin^{3} t}{3} + \frac{2}{3} \int \sin t \cdot (1 - \cos^{2} t) d t \\ = 2 t \sin t + 2 \cos t - \frac{2 t \sin^{3} t}{3} + \frac{2}{3} [\int \sin t d t - \int \sin t \cos^{2} t d t] \\ = 2 t \sin t + 2 \cos t - \frac{2 t \sin^{3} t}{3} + \frac{2}{3} [- \cos t + \frac{\cos^{3} t}{3} + c] \\ = 2 t \sin t + 2 \cos t - \frac{2 t \sin^{3} t}{3} + \frac{2}{3} [- \cos t] + \frac{2}{3} [\frac{\cos^{3} t}{3}] + c \\ = 2 \sqrt{x} \sin \sqrt{x} + \frac{4}{3} \cos \sqrt{x} - \frac{2 \sqrt{x} \sin^{3} \sqrt{x}}{3} + \frac{2}{9} [\cos^{3} \sqrt{x}] + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 57

Answer: $\frac{x \sin 3 x}{12} + \frac{\cos 3 x}{36} + \frac{3 x \sin x}{4} + \frac{3 \cos x}{4} + c$
Hint: $\cos^{3} x = \frac{\cos 3 x + 3 \cos x}{4}$
Given: $\int x \cos^{3} x d x$
Solution:
$∴ \cos^{3} x = \frac{\cos 3 x + 3 \cos x}{4}$
$∴ \int x \cos^{3} x d x = \int x [\frac{\cos 3 x + 3 \cos x}{4}] d x = \frac{1}{4} \int x \cos 3 x d x + \frac{3}{4} \int x \cos x d x$
$\begin{aligned} \Rightarrow \frac{1}{4} [x \int \cos 3 x d x - \int [\frac{d x}{d x} (\int \cos 3 x d x)] d x] + \frac{3}{4} [x \int \cos x d x - \int [\frac{d x}{d x} \int \cos x d x] d x] \\ \Rightarrow \frac{1}{4} [\frac{x \sin 3 x}{3} - \int \frac{\sin 3 x}{3} d x] + \frac{3}{4} (x \sin x - \int \sin x d x) \\ \Rightarrow \frac{1}{4} [\frac{x \sin 3 x}{3} + \frac{\cos 3 x}{9}] + \frac{3}{4} (x \sin x + \cos x) + c \\ = \frac{x \sin 3 x}{12} + \frac{\cos 3 x}{36} + \frac{3 x \sin x}{4} + \frac{3 \cos x}{4} + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 58

Answer: $\frac{1}{2} (x \cos^{- 1} x - \sqrt{1 - x^{2}}) + c$
Hint: Let $x = \cos Θ$
Given: $\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x$
Solution: Let

x = \cos Θ

d x = - \sin Θ d Θ

∴ \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x = \int \tan^{- 1} \sqrt{\frac{1 - \cos Θ}{1 + \cos Θ}} (- \sin Θ d Θ)

\begin{aligned} = - \int \tan^{- 1} \sqrt{\frac{2 \sin^{2} \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}} \sin θ d θ \\ = - \int \tan^{- 1} \tan \frac{θ}{2} \sin θ d θ \\ = - \frac{1}{2} \int θ \sin θ d θ \\ = - \frac{1}{2} [θ (- \cos θ) - \int 1 (- \cos θ) d θ] \end{aligned}

Applying by parts

= \frac{1}{2} [- Θ \cos Θ + \sin Θ]

\begin{aligned} = \frac{1}{2} θ \cos θ - \frac{1}{2} \sin θ \\ = \frac{1}{2} \cos^{- 1} x x - \frac{1}{2} \sqrt{1 - x^{2}} + c \\ = \frac{x}{2} \cos^{- 1} x - \frac{1}{2} \sqrt{1 - x^{2}} + c \\ = \frac{1}{2} (x \cos^{- 1} x - \sqrt{1 - x^{2}}) + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 59

Answer: $a [(\frac{x}{a} + 1) \tan^{- 1} \sqrt{\frac{x}{a}} - \sqrt{\frac{x}{a}}] + c$
Hint: Put

x = a \tan^{2} Θ

Given: $\int \sin^{- 1} (\sqrt{\frac{x}{a + x}}) d x$
Solution: $\int \sin^{- 1} (\sqrt{\frac{x}{a + x}}) d x$
Put $x = a \tan^{2} θ \Rightarrow d x = 2 a \tan θ \sec^{2} θ d θ$
$\int \sin^{- 1} (\sqrt{\frac{x}{a + x}}) d x = \int \sin^{- 1} (\sqrt{\frac{a \tan^{2} θ}{a + a \tan^{2} θ}}) 2 a \tan θ \sec^{2} θ d θ$
$\begin{aligned} = 2 a \int \sin^{- 1} (\sin θ) \tan θ \sec^{2} θ d θ \\ = 2 a \int θ \tan θ \sec^{2} θ d θ \end{aligned}$
Let

\tan Θ = t \Rightarrow \sec^{2} Θ d Θ = d t

\begin{aligned} \Rightarrow 2 a \int t \tan^{- 1} t d t \\ \Rightarrow 2 a [\tan^{- 1} t \int t d t - \int (\frac{1}{1 + t^{2}}) \frac{t^{2}}{2} d t] \\ \Rightarrow 2 a [\frac{t^{2}}{2} \tan^{- 1} t - \frac{1}{2} \int (1 - \frac{1}{1 + t^{2}}) d t] \\ \Rightarrow a [t^{2} \tan^{- 1} t - t + \sin^{- 1} t] + c \\ \Rightarrow a [\tan^{2} θ \tan θ - \tan θ + \tan θ] + c \\ \Rightarrow a [\frac{x}{a} \sqrt{\frac{x}{a}} - \sqrt{\frac{x}{a}} + \sqrt{\frac{x}{a}}] + c \end{aligned}

Indefinite Integrals Exercise 18.25 Question 60

Answer: $\frac{1}{2} [- \sin^{- 1} {(x)}^{2} \sqrt{(1 - x^{4}) + x^{2}}] + c$
Hint: Let $\sin^{- 1} (x^{2}) = t$
Given: $\int \frac{x^{3} \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x$
Solution: Let

\sin^{- 1} (x^{2}) = t \Rightarrow \frac{2 x}{\sqrt{1 - x^{4}}} d x = d t

$\int \frac{x^{3} \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x$
$\begin{aligned} = \frac{1}{2} \int \frac{2 x \sin^{- 1} (x^{2}) \times x^{2}}{\sqrt{1 - x^{4}}} d x \\ = \frac{1}{2} \int t \sin t d t \\ = \int [t \int \sin t d t + \int \cos t d t] + c \\ = \frac{1}{2} [- t \cos t + \sin t] + c \\ = \frac{1}{2} [- \sin^{- 1} (x)^{2} \sqrt{(1 - x^{4})} + x^{2}] + c \end{aligned}$

Indefinite Integrals Exercise 18.25 Question 61

Answer: $\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + l o g (\sqrt{1 - x^{2}}) - \frac{{(\sin^{- 1} x)}^{2}}{2} + c$
Hint:Put

\sin^{- 1} x = u

Given: $\int \frac{x^{2} \sin^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x$
Solution: Put

\sin^{- 1} x = u

\Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d u

and

x = \sin u

\begin{aligned} ∴ \int \frac{x^{2} \sin^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x = \int \frac{x^{2} \sin^{- 1} x}{(1 - x^{2}) \sqrt{1 - x^{2}}} d x \\ \Rightarrow \int (\frac{u \sin^{2} u}{1 - \sin^{2} u}) d u \\ = \int \frac{u \sin^{2} u}{\cos^{2} u} d u = \int u \tan^{2} u d u = \int u (\sec^{2} u - 1) d u \end{aligned}

\begin{aligned} = \int u \sec^{2} u d u - \int u d u \\ = u \times \int \sec^{2} u d u - \int \frac{d}{d u} u \int \sec^{2} u d u d u - \frac{u^{2}}{2} + c \\ = u \tan u - \int 1 \times \tan u d u - \frac{u^{2}}{2} + c \\ = u \tan u - \int \frac{\sin u}{\cos u} d u - \frac{u^{2}}{2} + c \end{aligned}

Put

\cos u = v \Rightarrow - \sin u d u = d v \Rightarrow \sin u d u = - d v

\begin{aligned} = u \tan u + \int \frac{d v}{v} - \frac{u^{2}}{2} + c \\ = u \tan u + \ln v - \frac{u^{2}}{2} + c \\ = u \frac{\sin u}{\cos u} + \ln (\sqrt{1 - \sin^{2} u}) - \frac{u^{2}}{2} + c \\ = u \frac{\sin u}{\sqrt{1 - \sin^{2} u}} + \ln (\sqrt{1 - \sin^{2} u}) - \frac{u^{2}}{2} + c \\ = \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \log (\sqrt{1 - x^{2}}) - \frac{{(\sin^{- 1} x)}^{2}}{2} + c \end{aligned}

RD Sharma class 12th exercise 18.25 contains 61 questions from the RD Sharma class 12 solution of Indefinite integrals exercise 18.25 that are presented in the most basic format and hence the concepts covered are -

Applications of integration
Applications of indefinite integration
Area under a curve by integration
Centroid of area by integration
Some integrals theorems

The advantages for using the RD Sharma Class 12th exercise 18.25 solution are listed below:-

The RD Sharma class 12 solution chapter 18 exercise 18.25 consists of a lot of questions that give a brief understanding of the chapter.
The solutions of the RD Sharma class 12th exercise 18.25 are designed by experts of maths and give you extraordinary questions to tighten your maths skills.
The solutions also provide helpful tips and solved questions for better reference and understanding maths in an easy way.
Teachers use the solution of RD Sharma class 12 chapter 18 exercise 18.25 for preparing lectures and giving homeworks.
The RD Sharma class 12th exercise 18.25 can be downloaded from Careers360 website for online PDFs and study materials for free of cost.
The solutions are trusted by every student, teacher, and schools as well for nurturing the students with the authentic knowledge of mathematics.

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RD Sharma Class 12 Exercise 18.25 Indefinite integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.25

Answer: logx(x22+x)−x24−x+c

The advantages for using the RD Sharma Class 12th exercise 18.25 solution are listed below:-

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Answer: $l o g x (\frac{x^{2}}{2} + x) - \frac{x^{2}}{4} - x + c$