NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

Edited By Komal Miglani | Updated on Apr 27, 2025 09:08 AM IST

Integration is considered an important pillar of calculus, and it has different applications in real life. You will use integration to find the area under the curves, volumes of irregular forms, and total cumulative quantities such as money or work completed. It is imperative in the field of physics to comprehend forces such as friction and gravity, calculate the centre of mass, and analyse motion. As we have studied all the integration concepts in this exercise, you will find many new and advanced questions that will give more clarity on the topic. The NCERT Solutions for the Miscellaneous Exercise of Chapter 7 of Class 12 are being prepared by the subject matter experts at Careers360 to provide a clear and appropriate approach to solve the questions. For syllabus, notes, and PDF, refer to this link: NCERT.

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NCERT solutions for class 12 maths chapter 7 Integrals- Miscellaneous Exercise

Question 1: Integrate the functions in Exercises 1 to 24.

$\frac{1}{x - x^{3}}$

Answer:

Firstly we will simplify the given equation :-

$\frac{1}{x - x^{3}} = \frac{1}{(x) (1 - x) (1 + x)}$

Let

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

By solving the equation and equating the coefficients of x 2 , x and the constant term, we get

$A = 1, B = \frac{1}{2}, C = \frac{- 1}{2}$

Thus the integral can be written as :

$\int \frac{1}{(x) (1 - x) (1 + x)} d x = \int \frac{1}{x} d x + \frac{1}{2} \int \frac{1}{1 - x} d x + \frac{- 1}{2} \int \frac{1}{1 + x} d x$

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

or $= \frac{1}{2} \log \frac{x^{2}}{1 - x^{2}} + C$

Question 2: Integrate the functions in Exercises 1 to 24.

$\frac{1}{\sqrt{x + a} + \sqrt{x + b}}$

Answer:

At first we will simplify the given expression,

$\frac{1}{\sqrt{x + a} + \sqrt{x + b}} = \frac{1}{\sqrt{x + a} + \sqrt{x + b}} \times \frac{\sqrt{x + a} - \sqrt{x + b}}{\sqrt{x + a} - \sqrt{x + b}}$

or $= \frac{\sqrt{x + a} - \sqrt{x + b}}{a - b}$

Now taking its integral we get,

$\int \frac{1}{\sqrt{x + a} + \sqrt{x + b}} = \frac{1}{a - b} \int (\sqrt{x + a} - \sqrt{x + b}) d x$

or $= \frac{1}{a - b} [\frac{(x + a)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{(x + b)^{\frac{3}{2}}}{\frac{3}{2}}]$

or $= \frac{2}{3 (a - b)} [(x + a)^{\frac{3}{2}} - (x + b)^{\frac{3}{2}}] + C$

Question 3: Integrate the functions in Exercises 1 to 24.

$\frac{1}{x \sqrt{a x - x^{2}}}$ [Hint: Put $x = \frac{a}{t}$ ]

Answer:

Let

$x = \frac{a}{t} d x \Rightarrow d x = \frac{- a}{t^{2}} d h$

Using the above substitution we can write the integral is

$\int \frac{1}{x \sqrt{a x - x^{2}}} = \int \frac{1}{\frac{a}{t} \sqrt{a . \frac{a}{t} - (\frac{a}{t})^{2}}} \frac{- a}{t^{2}} d t$

$= \frac{- 1}{a} \int \frac{1}{\sqrt{(t - 1)}} d t$

$= \frac{- 1}{a} (2 \sqrt{t - 1}) + C$

or $= \frac{- 1}{a} (2 \sqrt{\frac{a}{x} - 1}) + C$

or $= \frac{- 2}{a} \sqrt{\frac{a - x}{x}} + C$

Question 4: Integrate the functions in Exercises 1 to 24.

. $\frac{1}{x^{2} (x^{4} + 1)^{\frac{3}{4}}}$

Answer:

For the simplifying the expression, we will multiply and dividing it by x -3 .

We then have,

$\frac{x^{- 3}}{x^{2} x^{- 3} (x^{4} + 1)^{\frac{3}{4}}} = \frac{1}{x^{5}} {[\frac{x^{4} + 1}{x^{4}}]}^{\frac{- 3}{4}}$

Now, let

$\frac{1}{x^{4}} = t \Rightarrow \frac{1}{x^{5}} d x = \frac{- d t}{4}$

Thus,

$\int \frac{1}{x^{2} (x^{4} + 1)^{\frac{3}{4}}} = \int \frac{1}{x^{5}} (1 + {\frac{1}{x^{4}}}^{\frac{- 3}{4}}) d x$

or $= \frac{- 1}{4} \int (1 + t)^{\frac{- 3}{4}} d t$

$= \frac{- 1}{4} \frac{(1 + \frac{1}{x^{4}})^{\frac{1}{4}}}{\frac{1}{4}} + C$

$= - {[1 + \frac{1}{x^{4}}]}^{\frac{1}{4}} + C$

Question 5: Integrate the functions in Exercises 1 to 24.

$\frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}$ [Hint: $\frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}} = \frac{1}{x^{\frac{1}{3}} (1 + x^{\frac{1}{6}})}$ , put $x = t^{6}$ ]

Answer:

Put $x = t^{6} \Rightarrow d x = 6 t^{5} d t$

We get,

$\int \frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}} d x = \int \frac{6 t^{5}}{t^{3} + t^{2}} d t$

or $= 6 \int \frac{t^{3}}{1 + t} d t$

or $= 6 \int {(t^{2} - t + 1) - \frac{1}{1 + t}} d t$

or $= 6 [(\frac{t^{3}}{3}) - (\frac{t^{2}}{2}) + t - \log (1 + t)]$

Now put $x = t^{6}$ in the above result :

$= 2 \sqrt{x} - 3 x^{\frac{1}{3}} + 6 x^{\frac{1}{6}} - 6 \log (1 - x^{\frac{1}{6}}) + C$

Question 6: Integrate the functions in Exercises 1 to 24.

$\frac{5 x}{(x + 1) (x^{2} + 9)}$

Answer:

Let us assume that :

$\frac{5 x}{(x + 1) (x^{2} + 9)} = \frac{A}{(x + 1)} + \frac{B x + c}{x^{2} + 9}$

Solving the equation and comparing coefficients of x 2 , x and the constant term.

We get,

$A = \frac{- 1}{2}; B = \frac{1}{2}; C = \frac{9}{2}$

Thus the equation becomes :

$\frac{5 x}{(x + 1) (x^{2} + 9)} = \frac{- 1}{2 (x + 1)} + \frac{\frac{x}{2} + \frac{9}{2}}{x^{2} + 9}$

$\int \frac{5 x}{(x + 1) (x^{2} + 9)} = \int [\frac{- 1}{2 (x + 1)} + \frac{x + 9}{2 (x^{2} + 9})] d x$

or $= \frac{- 1}{2} \log | x + 1 | + \frac{1}{2} \int \frac{x}{x^{2} + 9} d x + \frac{9}{2} \int \frac{1}{x^{2} + 9} d x$

or $= \frac{- 1}{2} \log | x + 1 | + \frac{1}{4} \int \frac{2 x}{x^{2} + 9} d x + \frac{9}{2} \int \frac{1}{x^{2} + 9} d x$

or $= \frac{- 1}{2} \log | x + 1 | + \frac{1}{4} \log (x^{2} + 9) + \frac{3}{2} \tan^{- 1} \frac{x}{3} + C$

Question 7: Integrate the functions in Exercises 1 to 24.

$\frac{\sin x}{\sin (x - a)}$

Answer:

We have,

$I = \frac{\sin x}{\sin (x - a)}$

Assume :- $(x - a) = t \Rightarrow d x = d t$

Putting this in above integral :

$\int \frac{\sin x}{\sin (x - a)} d x = \int \frac{\sin (t + a)}{\sin t} d t$

or $= \int \frac{\sin t \cos a + \cos t \sin a}{\sin t} d t$

or $= \int (\cos a + \cot t \sin a) d t$

or $= t \cos a + \sin a \log | \sin t | + C$

or $= \sin a \log | \sin (x - a) | + x \cos a + C$

Question 8: Integrate the functions in Exercises 1 to 24.

$\frac{\cos x}{\sqrt{4 - \sin^{2} x}}$

Answer:

We have the given integral

$I = \frac{\cos x}{\sqrt{4 - \sin^{2} x}}$

Assume $\sin x = t \Rightarrow \cos x d x = d t$

So, this substitution gives,

$\int \frac{\cos x}{\sqrt{4 - \sin^{2} x}} = \int \frac{d t}{\sqrt{(2)^{2} - (t)^{2}}}$

$= \sin^{- 1} \frac{t}{2} + C$

or $= \sin^{- 1} (\frac{\sin x}{2}) + C$

Question 10: Integrate the functions in Exercises 1 to 24.

$\frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{x} \cos^{2} x}$

Answer:

We have

$I = \int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{x} \cos^{2} x}$

Simplifying the given expression, we get :

$\frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{x} \cos^{2} x} = \frac{(\sin^{4} x + \cos^{4} x) (\sin^{4} x - \cos^{4} x)}{1 - 2 \sin^{x} \cos^{2} x}$

or $= \frac{(\sin^{4} x + \cos^{4} x) (\sin^{2} x - \cos^{2} x) (\sin^{2} x + \cos^{2} x)}{1 - 2 \sin^{x} \cos^{2} x}$

or $= - \frac{(\sin^{4} x + \cos^{4} x) (\cos^{2} x - \sin^{2} x)}{1 - 2 \sin^{x} \cos^{2} x}$

or $= - \cos^{2} x - \sin^{2} x = - \cos 2 x$

Thus,

$I = \int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{x} \cos^{2} x} = - \int \cos 2 x d x$

and $= - \frac{\sin 2 x}{2} + C$

Question 11: Integrate the functions in Exercises 1 to 24.

$\frac{1}{\cos (x + a) \cos (x + b)}$

Answer:

For simplifying the given equation, we need to multiply and divide the expression by $\sin (a - b)$ .

Thus we obtain :

$\frac{1}{\cos (x + a) \cos (x + b)} = \frac{1}{\sin (a - b)} \times \frac{\sin (a - b)}{\cos (x + a) \cos (x + b)}$

or $= \frac{1}{\sin (a - b)} \times \frac{\sin [(x + a) - (x + b)]}{\cos (x + a) \cos (x + b)}$

or $= \frac{1}{\sin (a - b)} \times (\frac{\sin (x + a)}{\cos (x + a)} - \frac{\sin (x + b)}{\cos (x + b)})$

or $= \frac{1}{\sin (a - b)} \times (\tan (x + a) - \tan (x + b))$

Thus integral becomes :

$\int \frac{1}{\cos (x + a) \cos (x + b)} = \frac{1}{\sin (a - b)} \times \int (\tan (x + a) - \tan (x + b)) d x$

or $= \frac{1}{\sin (a - b)} \times [- \log | \cos (x + a) | + \log | \cos (x + b) |] + C$

or $= \frac{1}{\sin (a - b)} \times \log [\frac{\cos (x + b)}{c o s (x + a)}] + C$

Question 12: Integrate the functions in Exercises 1 to 24.

$\frac{x^{3}}{\sqrt{1 - x^{8}}}$

Answer:

Given that to integrate

$\frac{x^{3}}{\sqrt{1 - x^{8}}}$

Let $x^{4} = t ⟹ 4 x^{3} d x = d t$

$∴ \int \frac{x^{3}}{\sqrt{1 - x^{8}}} d x = \frac{1}{4} \int \frac{1}{\sqrt{1 - t^{2}}} d t$

$= \frac{1}{4} s i n^{- 1} t + C = \frac{1}{4} s i n^{- 1} x^{4} + C$

the required solution is $\frac{1}{4} s i n^{- 1} (x^{4}) + C$

Question 13: Integrate the functions in Exercises 1 to 24.

$\frac{e^{x}}{(1 + e^{x}) (2 + e^{x})}$

Answer:

we have to integrate the following function

$\frac{e^{x}}{(1 + e^{x}) (2 + e^{x})}$

Let $1 + e^{x} = t ⟹ e^{x} d x = d t$

using this we can write the integral as

$∴ \int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})} d x = \int \frac{1}{t (1 + t)} d t = \int \frac{(1 + t) - t}{t (1 + t)} d t$

$= \int (\frac{1}{t} - \frac{1}{t + 1}) d t$

$= \int \frac{1}{t} d t - \int \frac{1}{t + 1} d t$

$= \log t - \log (1 + t) + C = \log (1 + e^{x}) - \log (2 + e^{x}) + C = \log (\frac{e^{x} + 1}{e^{x} + 2}) + C$

Question 14: Integrate the functions in Exercises 1 to 24.

$\frac{1}{(x^{2} + 1) (x^{2} + 4)}$

Answer:

Given,

$\frac{1}{(x^{2} + 1) (x^{2} + 4)}$

Let $I = \int \frac{1}{(x^{2} + 1) (x^{2} + 4)}$

Now, Using partial differentiation,

$\frac{1}{(x^{2} + 1) (x^{2} + 4)} = \frac{A x + B}{(x^{2} + 1)} + \frac{C x + D}{(x^{2} + 4)}$

$⟹ \frac{1}{(x^{2} + 1) (x^{2} + 4)} = \frac{(A x + B) (x^{2} + 4) + (C x + D) (x^{2} + 1)}{(x^{2} + 1) (x^{2} + 4)}$
$⟹ 1 = (A x + B) (x^{2} + 4) + (C x + D) (x^{2} + 1) ⟹ 1 = A x^{3} + 4 A x + B x^{2} + 4 B + C x^{3} + C x + D x^{2} + D ⟹ (A + C) x^{3} + (B + D) x^{2} + (4 A + C) x + (4 B + D) = 1$

Equating the coefficients of $x, x^{2}, x^{3}$ and constant value,

A + C = 0 $⟹$ C = -A

B + D = 0 $⟹$ B = -D

4A + C =0 $⟹$ 4A = -C $⟹$ 4A = A $⟹$ A = 0 = C

4B + D = 1 $⟹$ 4B – B = 1 $⟹$ B = 1/3 = -D

Putting these values in equation, we have

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$⟹ I = \frac{1}{3} t a n^{- 1} x - \frac{1}{6} t a n^{- 1} \frac{x}{2} + C$

Question 15: Integrate the functions in Exercises 1 to 24.

$\cos^{3} x e^{\log \sin x}$

Answer:

Given,

$\cos^{3} x e^{\log \sin x}$

$I = \int \cos^{3} x e^{\log \sin x}$ (let)

Let $c o s x = t ⟹ - s i n x d x = d t ⟹ s i n x d x = - d t$

using the above substitution the integral is written as

$∴ \int c o s^{3} x e^{\log s i n x} d x = \int c o s^{3} x . s i n x d x$

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$I = - \frac{c o s^{4} x}{4} + C$

Question 16: Integrate the functions in Exercises 1 to 24.

$e^{3 \log x} (x^{4} + 1)^{- 1}$

Answer:

Given the function to be integrated as

$e^{3 \log x} (x^{4} + 1)^{- 1}$
$= e^{\log x^{3}} (x^{4} + 1)^{- 1} = \frac{x^{3}}{x^{4} + 1}$

Let $I = \int e^{3 \log x} (x^{4} + 1)^{- 1}$

Let $x^{4} = t ⟹ 4 x^{3} d x = d t$

$I = \int e^{3 \log x} (x^{4} + 1)^{- 1} = \int \frac{x^{3}}{x^{4} + 1}$

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$⟹ I = \frac{1}{4} \log (x^{4} + 1) + C$

Question 17: Integrate the functions in Exercises 1 to 24.

$f^{'} (a x + b) [f (a x + b)]^{n}$

Answer:

Given,

$f^{'} (a x + b) [f (a x + b)]^{n}$

Let $I = \int f^{'} (a x + b) [f (a x + b)]^{n}$

Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt

Now we can write the ntegral as

$\int f^{'} (a x + b) [f (a x + b)]^{n} = \frac{1}{a} \int t^{n} d t$

$= \frac{1}{a} . \frac{t^{n + 1}}{n + 1} + C = \frac{1}{a} . \frac{(f (a x + b))^{n + 1}}{n + 1} + C$

$⟹ I = \frac{(f (a x + b))^{n + 1}}{a (n + 1)} + C$

Question 18: Integrate the functions in Exercises 1 to 24.

. $\frac{1}{\sqrt{\sin^{3} x \sin (x + α)}}$

Answer:

Given,

$\frac{1}{\sqrt{\sin^{3} x \sin (x + α)}}$

Let $I = \int \frac{1}{\sqrt{\sin^{3} x \sin (x + α)}}$

We know the identity that

sin (A+B) = sin A cos B + cos A sin B

$∴ \frac{1}{\sqrt{\sin^{3} x \sin (x + α)}} = \frac{1}{\sqrt{\sin^{3} x (\sin x \cos α + \cos x \sin α)}}$

$= \frac{1}{\sqrt{\sin^{3} x . \sin x (\cos α + \cot x \sin α)}} = \frac{1}{\sqrt{\sin^{4} x (\cos α + \cot x \sin α)}}$

$\frac{c o s e c^{2} x}{\sqrt{(\cos α + \cot x \sin α)}}$

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Question 19: Integrate the functions in Exercises 1 to 24.

. $\frac{\sin^{- 1} \sqrt{x} - \cos^{- 1} \sqrt{x}}{\sin^{- 1} \sqrt{x} + \cos^{- 1} \sqrt{x}}, x \in [0, 1]$

Answer:

We have

$I = \int \frac{\sin^{- 1} \sqrt{x} - \cos^{- 1} \sqrt{x}}{\sin^{- 1} \sqrt{x} + \cos^{- 1} \sqrt{x}} d x$

or $= \int \frac{\sin^{- 1} \sqrt{x} - (\frac{Π}{2} - \sin^{- 1} \sqrt{x})}{\frac{Π}{2}} d x$

or $= \frac{2}{Π} \int (2 s i n^{- 1} \sqrt{x} - \frac{Π}{2}) d x$

or $= \int (\frac{4}{Π} \sin^{- 1} \sqrt{x} - 1) d x$

or $= \frac{4}{Π} \int \sin^{- 1} \sqrt{x} - 1 d x - \int 1 d x + C$

or $= \frac{4}{Π} \int \sin^{- 1} \sqrt{x} d x - x + C$

Thus $I = \frac{4}{Π} I^{'} - x + C$

Now we will solve I'.

$I^{'} = \int \sin^{- 1} \sqrt{x} d x$

Put x = t 2 .

Differentiating the equation wrt x, we get

$d x = 2 t d t$

Thus $\int \sin^{- 1} \sqrt{x} d x = \int \sin^{- 1} t 2 t d t$

or $= 2 \int t \sin^{- 1} t d t$

Using integration by parts, we get :

$= 2 [\sin^{- 1} t \int t d t - \int ((\frac{d}{d t} \sin^{- 1} t) \int t d t)] d t$

or $= t^{2} \sin^{- 1} t - \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t + C^{'}$

We know that

$\int \frac{- t^{2}}{\sqrt{1 - t^{2}}} d t = \frac{t}{2} \sqrt{1 - t^{2}} - \frac{1}{2} \sin^{- 1} t$

Thus it becomes :

$I^{'} = t^{2} \sin^{- 1} t + \frac{t}{2} \sqrt{1 - t^{2}} - \frac{1}{2} \sin^{- 1} t$

So I come to be :-

$I = \frac{4}{Π} I^{'} - x + C$

$I = \sin^{- 1} \sqrt{x} [\frac{2 (2 x - 1)}{Π}] + \frac{2 \sqrt{x - x^{2}}}{Π} - x + C$

Question 20: Integrate the functions in Exercises 1 to 24.

$\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}}$

Answer:

Given,

$\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}}$ = I (let)

Let $x = c o s^{2} θ ⟹ d x = - 2 s i n θ c o s θ d θ$

$A n d \sqrt{x} = c o s θ ⟹ θ = \cos^{- 1} \sqrt{x}$

using the above substitution we can write the integral as

$I = \int \sqrt{\frac{1 - \sqrt{c o s^{2} θ}}{1 + \sqrt{c o s^{2} θ}}} (- 2 \sin θ \cos θ) d θ = - \int \sqrt{\frac{1 - c o s θ}{1 + c o s θ}} (2 \sin θ \cos θ) d θ$

$= - \int \sqrt{t a n^{2} \frac{θ}{2}} (2 \sin θ \cos θ) d θ = - \int \sqrt{t a n^{2} \frac{θ}{2}} (2. 2 \sin \frac{θ}{2} \cos \frac{θ}{2} \cos θ) d θ = - 4 \int \sin^{2} \frac{θ}{2} \cos θ d θ$

$= - 4 \int \sin^{2} \frac{θ}{2} (2 c o s^{2} \frac{θ}{2} - 1) d θ$

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Question 21: Integrate the functions in Exercises 1 to 24.

$\frac{2 + \sin 2 x}{1 + \cos 2 x} e^{x}$

Answer:

Given to evaluate

$\frac{2 + \sin 2 x}{1 + \cos 2 x} e^{x}$

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now the integral becomes

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Let tan x = f(x)

$⟹ f^{'} (x) = s e c^{2} x d x$

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Question 22: Integrate the functions in Exercises 1 to 24.

$\frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)}$

Answer:

Given,

$\frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)}$

using partial fraction we can simplify the integral as

Let $\frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{x + 2}$

$⟹ \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} = \frac{A (x + 1) (x + 2) + B (x + 2) + C (x + 1)^{2}}{(x + 1)^{2} (x + 2)} ⟹ \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} = \frac{A (x^{2} + 3 x + 2) + B (x + 2) + C (x^{2} + 2 x + 1)}{(x + 1)^{2} (x + 2)}$

$⟹ x^{2} + x + 1 = A (x^{2} + 3 x + 2) + B (x + 2) + C (x^{2} + 2 x + 1) = (A + C) x^{2} + (3 A + B + 2 C) x + (2 A + 2 B + C)$

Equating the coefficients of x, x 2 and constant value, we get:

A + C = 1

3A + B + 2C = 1

2A+2B+C =1

Solving these:

A= -2, B=1 and C=3

$⟹ \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} = \frac{- 2}{x + 1} + \frac{1}{(x + 1)^{2}} + \frac{3}{x + 2}$

$⟹ \int \frac{x^{2} + x + 1}{(x + 1)^{2} (x + 2)} = \int \frac{- 2}{x + 1} d x + \int \frac{1}{(x + 1)^{2}} d x + \int \frac{3}{x + 2} d x = - 2 \log (x + 1) - \frac{1}{(x + 1)} + 3 \log (x + 2) + C$

Question 23: Integrate the functions in Exercises 1 to 24.

$\tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}$

Answer:

We have

$I = \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}$

Let us assume that : $x = \cos 2 Θ$

Differentiating wrt x,

$d x = - 2 \sin 2 Θ d Θ$

Substituting this in the original equation, we get

$\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} = \int \tan^{- 1} \sqrt{\frac{1 - c o s 2 Θ}{1 + c o s 2 Θ}} \times - 2 \sin 2 Θ d Θ$

or $= - 2 \int \tan^{- 1} (\frac{s i n Θ}{c o s Θ}) \times \sin 2 Θ d Θ$

or $= - 2 \int Θ \sin 2 Θ d Θ$

Using integration by parts , we get

$= - 2 (Θ \int \sin 2 Θ d Θ - \int \frac{d Θ}{d Θ} \int \sin 2 Θ d Θ)$

or $= - 2 (Θ (\frac{- \cos 2 Θ}{2}) - \int 1. \frac{- \cos 2 Θ}{2} d Θ)$

or $= - 2 (\frac{- Θ \cos 2 Θ}{2} + \frac{\sin 2 Θ}{4})$

Putting all the assumed values back in the expression,

$= - 2 (- \frac{1}{2} (\frac{1}{2} \cos^{- 1} x) + \frac{\sqrt{1 - x^{2}}}{4})$

or $= \frac{1}{2} (x \cos^{- 1} x - \sqrt{1 - x^{2}}) + C$

Question 24: Integrate the functions in Exercises 1 to 24.

$\frac{\sqrt{x^{2} + 1} [\log (x^{2} + 1) - 2 \log x]}{x^{4}}$

Answer:

$\frac{\sqrt{x^{2} + 1} [\log (x^{2} + 1) - 2 \log x]}{x^{4}}$

Here let's first reduce the log function.

$= \frac{\sqrt{x^{2} + 1}}{x^{4}} [\log (x^{2} + 1) - \log x^{2}] d x$

$= \frac{\sqrt{x^{2} (1 + \frac{1}{x^{2}})}}{x^{4}} [\log \frac{(x^{2} + 1)}{x^{2}}] d x$

$= \int \frac{\sqrt{(1 + \frac{1}{x^{2}})}}{x^{3}} [\log (1 + \frac{1}{x^{2}})] d x$

Now, let

$t = 1 + \frac{1}{x^{2}}$

$d t = \frac{- 2}{x^{3}} d x$

So our function in terms if new variable t is :

$I = \frac{- 1}{2} \int [\log t] \cdot t^{\frac{1}{2}} d t$

now let's solve this By using integration by parts

$I = \frac{- 1}{2} \int [(\log t) \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - \int \frac{1}{t} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} d t]$

$I = \frac{- 1}{3} t^{\frac{3}{2}} \log t + \frac{1}{3} \int t^{\frac{1}{2}} d t$

$I = \frac{- 1}{3} t^{\frac{3}{2}} \log t + \frac{1}{3} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}$

$I = \frac{2}{9} t^{\frac{3}{2}} - \frac{1}{3} t^{\frac{3}{2}} l o g t + c$

$I = \frac{1}{3} t^{\frac{3}{2}} [\frac{2}{3} - \log t] + c$

$I = \frac{1}{3} {(1 + \frac{1}{x^{2}})}^{\frac{3}{2}} [\frac{2}{3} - \log (1 + \frac{1}{x^{2}})] + c$

Question 25: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x$

Answer:

$\int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x$

Since, we have $e^{x}$ multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

$\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x)$

So,

$\int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x$

$= \int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - 2 \sin \frac{x}{2} c o s \frac{x}{2}}{2 s i n^{2} \frac{x}{2}}) d x$

$= \int_{\frac{π}{2}}^{π} e^{x} (\frac{1}{2 s i n^{2} \frac{x}{2}} - \frac{2 \sin \frac{x}{2} c o s \frac{x}{2}}{2 s i n^{2} \frac{x}{2}}) d x$

$= \int_{\frac{π}{2}}^{π} e^{x} (\frac{1}{2} c o s e c^{2} \frac{x}{2} - c o t \frac{x}{2}) d x$

$= \int_{\frac{π}{2}}^{π} e^{x} (- c o t \frac{x}{2} + \frac{1}{2} c o s e c^{2} \frac{x}{2}) d x$

Here let's use the property

$\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x)$

so,

$= \int_{\frac{π}{2}}^{π} e^{x} (- c o t \frac{x}{2} + \frac{1}{2} c o s e c^{2} \frac{x}{2}) d x$

$= {[- e^{x} c o t \frac{x}{2}]}_{\frac{π}{2}}^{π}$

$= [- e^{π} c o t \frac{π}{2}] - [- e^{\frac{π}{2}} c o t \frac{π}{4}]$

$= e^{\frac{π}{2}}$

Question 26: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{0}^{\frac{π}{4}} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x}$

Answer:

$\int_{0}^{\frac{π}{4}} \frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x}$

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let' divide both numerator and denominator by $c o s^{4} x$

$= \int_{0}^{\frac{π}{4}} \frac{\frac{\sin x \cos x}{c o s x c o s x s o s^{2} x}}{1 + \frac{\sin^{4} x}{\cos^{4} x}}$

$= \int_{0}^{\frac{π}{4}} \frac{t a n x s e c^{2} x}{1 + t a n^{4} x}$

Now lets change the variable

$t = t a n^{2} x d t = 2 t a n x s e c^{2} x d x$

the limits will also change since the variable is changing

$w h e n x = 0, t = t a n^{2} 0 = 0$

$w h e n x = \frac{π}{4}, t = t a n^{2} \frac{π}{4} = 1$

So, the integration becomes:

$I = \frac{1}{2} \int_{0}^{1} \frac{d t}{1 + t^{2}}$

$I = \frac{1}{2} {[t a n^{- 1} t]}_{0}^{1}$

$I = \frac{1}{2} [t a n^{- 1} 1] - \frac{1}{2} [t a n^{- 1} 0]$

$I = \frac{1}{2} [\frac{π}{4}] - 0$

$I = \frac{π}{8}$

Question 27: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{0}^{\frac{π}{2}} \frac{\cos^{2} x d x}{\cos^{2} x + 4 \sin^{2} x}$

Answer:

Lets first simplify the function.

$\int_{0}^{\frac{π}{2}} \frac{\cos^{2} x d x}{\cos^{2} x + 4 \sin^{2} x} = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x d x}{\cos^{2} x + 4 (1 - \cos^{2} x)} = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x d x}{4 - 3 \cos^{2} x}$

$\frac{- 1}{3} \int_{0}^{\frac{π}{2}} \frac{4 - 3 \cos^{2} x - 4}{4 - 3 \cos^{2} x} d x = \frac{- 1}{3} \int_{0}^{\frac{π}{2}} \frac{4 - 3 \cos^{2} x}{4 - 3 \cos^{2} x} d x - \frac{- 1}{3} \int_{0}^{\frac{π}{2}} \frac{- 4}{4 - 3 \cos^{2} x} d x$

$= \frac{- 1}{3} \int_{0}^{\frac{π}{2}} 1 d x - \frac{- 1}{3} \int_{0}^{\frac{π}{2}} \frac{- 4}{4 - 3 \cos^{2} x} d x = \frac{- 1}{3} {[x]}_{0}^{\frac{π}{2}} - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4}{4 - 3 \cos^{2} x} d x$

As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

$= \frac{- 1}{3} [\frac{π}{2 - 0}] - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4 s e c^{2} x}{4 s e c^{2} x - 3} d x$

AS we can write square of sec in term of tan,

$= \frac{- 1}{3} [\frac{π}{2 - 0}] - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4 s e c^{2} x}{4 (1 + t a n^{2} x) - 3} d x$

$= \frac{- 1}{3} [\frac{π}{2 - 0}] - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4 s e c^{2} x}{1 + 4 t a n^{2} x} d x$

Now let's calculate the integral of the second function, (we already have calculated the first function)

$= - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4 s e c^{2} x}{1 + 4 t a n^{2} x} d x$

let

$t = 2 t a n x, d t = 2 s e c^{2} x d x$

here we are changing the variable so we have to calculate the limits of the new variable

when x = 0, t = 2tanx = 2tan(0)=0

when $x = π / 2, t = 2 t a n π / 2 = \infty$

our function in terms of t is

$= - \frac{2}{3} \int_{0}^{\infty} \frac{1}{1 + t^{2}} d t$

$= {[t a n^{- 1} t]}_{0}^{\infty} = [t a n^{- 1} \infty - t a n^{- 1} 0]$

$= \frac{π}{2}$

Hence our total solution of the function is

$= - \frac{π}{6} + \frac{2}{3} * \frac{π}{2} = \frac{π}{6}$

Question 28: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2 x}}$

Answer:

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2 x}}$

Here first let convert sin2x as the angle of x ( sinx, and cosx)

$= \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin x + \cos x}{\sqrt{2 s i n x c o s x}}$

Now let's remove the square root form function by making a perfect square inside the square root

$= \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin x + \cos x}{\sqrt{- (- 1 + 1 - 2 s i n x c o s x)}}$

$= \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin x + \cos x}{\sqrt{(1 - (s i n^{2} x + c o s^{2} x - 2 s i n x c o s x)}}$

$= \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sin x + \cos x}{\sqrt{(1 - (s i n x - c o s x)^{2}}}$

Now let

, $t = s i n x - c o s x d t = (c o s x + s i n x) d x$

since we are changing the variable, limit of integration will change

$w h e n x = π / 6, t = s i n π / 6 - c o s π / 6 = (1 - \sqrt{3}) / 2 w h e n x = π / 3, t = s i n π / 3 - c o s / p i / 3 = (\sqrt{3} - 1) / 2$

our function in terms of t :

$= \int_{\frac{1 - \sqrt{3}}{2}}^{\frac{\sqrt{3} - 1}{2}} \frac{1}{\sqrt{(1 - t^{2})}} d t$

$= {[s i n^{- 1} t]}_{\frac{1 - \sqrt{3}}{2}}^{\frac{\sqrt{3} - 1}{2}} = 2 s i n^{- 1} (\frac{\sqrt{3} - 1}{2})$

Question 29: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{0}^{1} \frac{d x}{\sqrt{1 + x} - \sqrt{x}}$

Answer:

$\int_{0}^{1} \frac{d x}{\sqrt{1 + x} - \sqrt{x}}$

First, let's get rid of the square roots from the denominator,

$= \int_{0}^{1} \frac{d x}{\sqrt{1 + x} - \sqrt{x}} * \frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}}$

$= \int_{0}^{1} \frac{\sqrt{1 + x} + \sqrt{x}}{1 + x - x} d x$

$= \int_{0}^{1} (\sqrt{1 + x} + \sqrt{x}) d x$

$\=\int_0^1({\sqrt{1+x})dx+\int_0^1({\sqrt{x}})dx$

$= \int_{0}^{1} (1 + x)^{\frac{1}{2}} d x + \int_{0}^{1} x^{\frac{1}{2}} d x$

$= {[\frac{2}{3} (1 + x)^{\frac{3}{2}}]}_{0}^{1} + {[\frac{2}{3} (x)^{\frac{3}{2}}]}_{0}^{1}$

$= [\frac{2}{3} (1 + 1)^{\frac{3}{2}}] - [\frac{2}{3}] + [\frac{2}{3} (1)^{\frac{3}{2}}] - [0]$

$= \frac{4 \sqrt{2}}{3}$

Question 30: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{9 + 16 \sin 2 x} d x$

Answer:

$\int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{9 + 16 \sin 2 x} d x$

First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt

So,

Now since we are changing the variable, the new limit of the integration will be,

when x = 0, t = cos0-sin0=1-0=1

when $x = π / 4$ $t = \cos π / 4 - \sin π / 4 = 0$

Now,

$(\cos x - \sin x)^{2} = t^{2}$

$\cos^{2} x + \sin^{2} x - 2 \cos x \sin x = t^{2}$

$1 - \sin 2 x = t^{2}$

$\sin 2 x = 1 - t^{2}$

Hence our function in terms of t becomes,

$\int_{- 1}^{0} \frac{d t}{9 + 16 (1 - t^{2})} = \int_{- 1}^{0} \frac{d t}{9 + 16 - 16 t^{2}} = \int_{- 1}^{0} \frac{d t}{25 - 16 t^{2}} = \int_{- 1}^{0} \frac{d t}{5^{2} - (4 t)^{2})}$

$= \frac{1}{4} {[\frac{1}{2 (5)} \log \frac{5 + 4 t}{5 - 4 t}]}_{- 1}^{0}$

$= \frac{1}{40} [\log (1) - \log (\frac{1}{9})]$

$= \frac{\log 9}{40}$

Question 31: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{0}^{\frac{π}{2}} \sin 2 x \tan^{- 1} (\sin x) d x$

Answer:

Let I =

$\int_{0}^{\frac{π}{2}} s i n 2 x t a n^{- 1} (s i n x) d x$

$= \int_{0}^{\frac{π}{2}} 2 s i n x c o s x t a n^{- 1} (s i n x) d x$

Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so

$t = s i n x d t = c o s x d x$

Now the important step here is to change the limit of the integration as we are changing the variable.so,

$w h e n x = 0, t = s i n 0 = 0 w h e n x = \frac{π}{2}, t = s i n \frac{π}{2} = 1$

So our function becomes,

$I = 2 \int_{0}^{1} (t a n^{- 1} t) t d t$

Now, let's integrate this by using integration by parts method,

$I = 2 {[t a n^{- 1} t \cdot \frac{t^{2}}{2} - \int \frac{1}{1 + t^{2}} \cdot \frac{t^{2}}{2} d t]}_{0}^{1}$

$I = 2 {[t a n^{- 1} t \cdot \frac{t^{2}}{2} - \frac{1}{2} \int \frac{t^{2}}{1 + t^{2}} \cdot d t]}_{0}^{1}$

$I = 2 {[t a n^{- 1} t \cdot \frac{t^{2}}{2} - \frac{1}{2} \int \frac{(1 + t^{2}) - 1}{1 + t^{2}} \cdot d t]}_{0}^{1}$

$I = 2 {[t a n^{- 1} t \cdot \frac{t^{2}}{2} - \frac{1}{2} \int (1 - \frac{1}{1 + t^{2}}) \cdot d t]}_{0}^{1}$

$I = 2 {[t a n^{- 1} t \cdot \frac{t^{2}}{2} - \frac{1}{2} (t - t a n^{- 1} t)]}_{0}^{1}$

$I = 2 {[t a n^{- 1} t \cdot \frac{t^{2}}{2} - \frac{1}{2} (t) + \frac{1}{2} t a n^{- 1} t)]}_{0}^{1}$

$I = 2 {[\frac{1}{2} (t a n^{- 1} t \cdot (t^{2} + 1) - t)]}_{0}^{1}$

$I = {[(t a n^{- 1} t \cdot (t^{2} + 1) - t)]}_{0}^{1}$

$I = [(t a n^{- 1} (1) \cdot (1^{2} + 1) - 1)] - [(t a n^{- 1} (0) \cdot (0^{2} + 1) - 0)]$ $I = 2 t a n^{- 1} 1 - 1 = 2 \times \frac{π}{4} - 1$

$I = \frac{π}{2} - 1$

Question 32: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{0}^{π} \frac{x \tan x}{\sec x + \tan x} d x$

Answer:

Let I = $\int_{0}^{π} \frac{x \tan x}{\sec x + \tan x} d x$ -(i)

Replacing x with ( $π$ -x),

$I = \int_{π}^{0} \frac{(π - x) \tan (π - x)}{\sec (π - x) + \tan (π - x)} (- d x) = - \int_{π}^{0} \frac{(π - x) (-) \tan x}{- \sec x - \tan x} d x$

$⟹ I = \int_{0}^{π} \frac{(π - x) \tan x}{\sec x + \tan x} d x$ - (ii)

Adding (i) and (ii)

$I + I = \int_{0}^{π} (\frac{x \tan x}{\sec x + \tan x} + \frac{(π - x) \tan x}{\sec x + \tan x}) d x$

$⟹ 2 I = \int_{0}^{π} \frac{π \tan x}{\sec x + \tan x} d x$

$⟹ 2 I = \int_{0}^{π} \frac{π \frac{s i n x}{c o s x}}{\frac{1}{c o s x} + \frac{s i n x}{c o s x}} d x ⟹ 2 I = π \int_{0}^{π} \frac{s i n x}{1 + s i n x} d x ⟹ 2 I = π \int_{0}^{π} \frac{(1 + s i n x) - 1}{1 + s i n x} d x ⟹ 2 I = π \int_{0}^{π} [1 - \frac{1}{1 + s i n x}] d x$

$⟹ 2 I = π \int_{0}^{π} [1 - \frac{1}{1 + s i n x}] d x ⟹ 2 I = π \int_{0}^{π} 1 d x - π \int_{0}^{π} \frac{1}{1 + s i n x} . \frac{(1 - s i n x)}{(1 - s i n x)} d x ⟹ 2 I = π \int_{0}^{π} 1 d x - π \int_{0}^{π} [\sec^{2} x - \sec x \tan x] d x ⟹ 2 I = π [x]_{0}^{π} - π [\sec x - \tan x]_{0}^{π}$

$⟹ 2 I = π [π - 0] - π [t a n π - s e c π - t a n π + s e c 0] ⟹ 2 I = π [π - 2] ⟹ I = \frac{π}{2} [π - 2]$

Question 33: Evaluate the definite integrals in Exercises 25 to 33.

$\int_{1}^{4} [| x - 1 | + | x - 2 | + | x - 3 |] d x$

Answer:

Given integral $\int_{1}^{4} [| x - 1 | + | x - 2 | + | x - 3 |] d x$

So, we split it in according to intervals they are positive or negative.

$= \int_{1}^{4} | x - 1 | d x + \int_{1}^{4} | x - 2 | d x + \int_{1}^{4} | x - 3 | d x$

$= I_{1} + I_{2} + I_{3}$

Now,

$I_{1} = \int_{1}^{4} | x - 1 | d x = \int_{1}^{4} (x - 1) d x$

$∵$ as $(x - 1)$ is positive in the given x -range $[1, 4]$

$= {[\frac{x^{2}}{2} - x]}_{1}^{4} = [\frac{4^{2}}{2} - 4] - [\frac{1^{2}}{2} - 1]$

$= [8 - 4] - [- \frac{1}{2}] = 4 + \frac{1}{2} = \frac{9}{2}$

Therefore, $I_{1} = \frac{9}{2}$

$I_{2} = \int_{1}^{4} | x - 2 | d x = \int_{1}^{2} (2 - x) d x + \int_{2}^{4} (x - 2) d x$

$∵$ as $(x - 2) \geq 0$ is in the given x -range $[2, 4]$ and $\leq 0$ in the range $[1, 2]$

$= {[2 x - \frac{x^{2}}{2}]}_{1}^{2} + {[\frac{x^{2}}{2} - 2 x]}_{2}^{4}$

$= {[2 (2) - \frac{2^{2}}{2}] - [2 (1) - \frac{1^{2}}{2}]} + {[\frac{4^{2}}{2} - 2 (4)] - [\frac{2^{2}}{2} - 2 (2)]}$

$= [4 - 2 - 2 + \frac{1}{2}] + [8 - 8 - 2 + 4]$

$= \frac{1}{2} + 2 = \frac{5}{2}$

Therefore, $I_{2} = \frac{5}{2}$

$I_{3} = \int_{1}^{4} | x - 3 | d x = \int_{1}^{3} (3 - x) d x + \int_{3}^{4} (x - 3) d x$

$∵$ as $(x - 3) \geq 0$ is in the given x -range $[3, 4]$ and $\leq 0$ in the range $[1, 3]$

$= {[3 x - \frac{x^{2}}{2}]}_{1}^{3} + {[\frac{x^{2}}{2} - 3 x]}_{3}^{4}$

$= {[3 (3) - \frac{3^{2}}{2}] - [3 (1) - \frac{1^{2}}{2}]} + {[\frac{4^{2}}{2} - 3 (4)] - [\frac{3^{2}}{2} - 3 (3)]}$

$= [9 - \frac{9}{2} - 3 + \frac{1}{2}] + [8 - 12 - \frac{9}{2} + 9]$

$= [6 - 4] + \frac{1}{2} = \frac{5}{2}$

Therefore, $I_{3} = \frac{5}{2}$

So, We have the sum $= I_{1} + I_{2} + I_{3}$

$I = \frac{9}{2} + \frac{5}{2} + \frac{5}{2} = \frac{19}{2}$

Question 34: Prove the following (Exercises 34 to 39)

. $\int_{1}^{3} \frac{d x}{x^{2} (x + 1)} = \frac{2}{3} + \log \frac{2}{3}$

Answer:

L.H.S = $\int_{1}^{3} \frac{d x}{x^{2} (x + 1)}$

We can write the numerator as [(x+1) -x]

$∴ \int_{1}^{3} \frac{d x}{x^{2} (x + 1)} = \int_{1}^{3} \frac{(x + 1) - x}{x^{2} (x + 1)} d x$

$= \int_{1}^{3} [\frac{1}{x^{2}} - \frac{1}{x (x + 1)}] d x = \int_{1}^{3} \frac{1}{x^{2}} d x - \int_{1}^{3} \frac{(x + 1) - x}{x (x + 1)} d x$

$= \int_{1}^{3} \frac{1}{x^{2}} d x - \int_{1}^{3} [\frac{1}{x} - \frac{1}{(x + 1)}] d x = \int_{1}^{3} \frac{1}{x^{2}} d x - \int_{1}^{3} \frac{1}{x} d x + \int_{1}^{3} \frac{1}{(x + 1)} d x = {[- \frac{1}{x}]}_{1}^{3} - {[\log x]}_{1}^{3} + {[\log (x + 1)]}_{1}^{3}$

$= [- \frac{1}{3} + 1] - [\log 3 - \log 1] + [\log 4 - \log 2] = \frac{2}{3} + \log (\frac{4}{3.2})$

$= \log (\frac{2}{3}) + \frac{2}{3}$ = RHS

Hence proved.

Question 35: Prove the following (Exercises 34 to 39)

$\int_{0}^{1} x e^{x} d x = 1$

Answer:

$L e t I = \int x e^{x} d x$

Integrating I by parts

$I = x \int e^{x} d x - \int ((\frac{d (x)}{d x}) \int e^{x} d x) d x I = x e^{x} - \int e^{x} d x I = x e^{x} - e^{x} + c$

Applying Limits from 0 to 1

$\int_{0}^{1} x e^{x} d x = [x e^{x} - e^{x} + c]_{0}^{1} I = [e - e + c] - [0 - 1 + c] I = 1$

Hence proved I = 1

Question 36: Prove the following (Exercises 34 to 39)

$\int_{- 1}^{1} x^{17} \cos^{4} x d x = 0$

Answer:

$L e t x^{17} c o s^{4} x = g (x)$

$g (- x) = (- x)^{17} c o s^{4} (- x) = - x^{17} c o s^{4} x = - g (x)$

The Integrand g(x) therefore is an odd function and therefore

$\int_{- 1}^{1} g (x) d x = 0$

Question 37: Prove the following (Exercises 34 to 39)

$\int_{0}^{\frac{π}{2}} \sin^{3} x d x = \frac{2}{3}$

Answer:

$L e t I = \int_{0}^{\frac{π}{2}} s i n^{3} x d x I = \int_{0}^{\frac{π}{2}} s i n x (1 - s i n^{2} x) d x I = \int_{0}^{\frac{π}{2}} s i n x d x - \int_{0}^{\frac{π}{2}} c o s^{2} x s i n x d x I = I_{1} - I_{2}$

$I_{1} = [- c o s x]_{0}^{\frac{π}{2}} I_{1} = - 0 - (- 1) = 1$

For I 2 let cosx=t, -sinxdx=dt

The limits change to 0 and 1

$I_{2} = - \int_{1}^{0} t^{2} d t I_{2} = - [\frac{t^{3}}{3}] {_{1}}^{0} I_{2} = 0 - (- \frac{1}{3}) I_{2} = \frac{1}{3}$

I 1 -I 2 =2/3

Hence proved.

Question 38: Prove the following (Exercises 34 to 39)

$\int_{0}^{\frac{π}{4}} 2 \tan^{3} x d x = 1 - \log 2$

Answer:

The integral is written as

$L e t I = \int 2 t a n^{3} x d x I = \int 2 t a n^{2} x \cdot t a n x d x I = \int 2 (s e c^{2} x - 1) t a n x d x I = 2 \int t a n x s e c^{2} x d x - 2 \int t a n x d x I = 2 \int t d t - 2 l o g (c o s x) + c (t = t a n x) I = t^{2} - 2 l o g (c o s x) + c I = t a n^{2} x - 2 l o g (c o s x) + c$

$[I]_{0}^{\frac{π}{4}} = [t a n^{2} x - 2 l o g (c o s x)]_{0}^{\frac{π}{4}}$

$[I]_{0}^{\frac{π}{4}} = (1 - 2 l o g \sqrt{2}) - (0 - 2 l o g 1)$

$[I]_{0}^{\frac{π}{4}} = 1 - l o g 2$

Hence Proved

Question 39: Prove the following (Exercises 34 to 39)\

$\int_{0}^{1} \sin^{- 1} x d x = \frac{π}{2} - 1$

Answer:

$L e t I = \int s i n^{-} x d x$

Integrating by parts we get

$I = s i n^{-} x \int 1 \cdot d x - \int (\frac{d (s i n^{-} x)}{d x} \int 1 \cdot d x) I = x s i n^{-} x + c - \int \frac{1}{\sqrt{1 - x^{2}}} \cdot x d x I = I_{1} - I_{2}$

For I 2 take 1-x 2 = t 2 , -xdx=tdt

$I_{2} = \int \frac{1}{\sqrt{1 - x^{2}}} \cdot x d x I_{2} = - \int \frac{1}{t} t d t I_{2} = - t + c I_{2} = - \sqrt{1 - x^{2}} + c$

$[I]_{0}^{1} = [I_{1} - I_{2}]_{0}^{1}$

$= [x s i n^{-} x - (- \sqrt{1 - x^{2}})]_{0}^{1} = [x s i n^{-} x + \sqrt{1 - x^{2}}]_{0}^{1} = [1 \cdot \frac{π}{2} + 0] - [0 + 1] = \frac{π}{2} - 1$

Hence Proved

Question 40: Evaluate $\int_{0}^{1} e^{2 - 3 x} d x$ as a limit of a sum.

Answer:

As we know

$\int_{a}^{b} f (x) d x = (b - a) lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . + f (a + (n - 1) h)]$

where b-a=hn

In the given problem b=1, a=0 and $f (x) = e^{2 - 3 x}$
$\int_{0}^{1} e^{2 - 3 x} d x = (1 - 0) lim_{n \to \infty} \frac{1}{n} (e^{2} + e^{2 - 3 h} + e^{2 - 3 (2 h)} . . . . . + e^{2 - 3 (n - 1) h}) = e^{2} lim_{n \to \infty} \frac{1}{n} (1 + e^{- 3 h} + e^{- 6 h} . . . . + e^{- 3 (n - 1) h}) = e^{2} lim_{n \to \infty} \frac{1}{n} (\frac{1 - (e^{- 3 h})^{n}}{1 - e^{- 3 h}}) = e^{2} lim_{n \to \infty} \frac{1}{n} (\frac{1 - e^{- \frac{3}{n} \times n}}{1 - e^{- \frac{3}{n}}})$

$= e^{2} lim_{n \to \infty} \frac{1}{n} (\frac{1 - e^{- 3}}{1 - e^{- \frac{3}{n}}}) = \frac{e^{2} (1 - e^{- 3})}{3} lim_{n \to \infty} \frac{- \frac{3}{n}}{e^{- \frac{3}{n}} - 1} = \frac{e^{2} (1 - e^{- 3})}{3}$

$= \frac{e^{2} - e^{- 1}}{3}$

Question 41: Choose the correct answers in Exercises 41 to 44.

. $\int \frac{d x}{e^{x} + e^{- x}}$ is equal to

(A) $\tan^{- 1} (e^{x}) + c$

(B) $\tan^{- 1} (e^{- x}) + c$

(D) $\log (e^{x} + e^{- x}) + C$

Answer:

$\int \frac{d x}{e^{x} + e^{- x}}$

the above integral can be re arranged as

$= \int \frac{e^{x}}{e^{2 x} + 1} d x$

let e x =t, e x dx=dt

$\int \frac{d x}{e^{x} + e^{- x}}$

$= \int \frac{1}{t^{2} + 1} d t = t a n^{- 1} t + c = t a n^{- 1} (e^{x}) + c$

(A) is correct

Question 42: Choose the correct answers in Exercises 41 to 44.

. $\int \frac{\cos 2 x}{(\sin x + \cos x)^{2}} d x$ is equal to

(A) $\frac{- 1}{\sin x + \cos x} + C$

(B) $\log | \sin x + \cos x | + C$

(D) $\frac{1}{(\sin x + \cos x)^{2}} + C$

Answer:

$\frac{\cos 2 x}{(\sin x + \cos x)^{2}} = \frac{c o s^{2} x - s i n^{2} x}{(\sin x + \cos x)^{2}} = \frac{(\sin x + \cos x) (\cos x - \sin x)}{(\sin x + \cos x)^{2}} = \frac{(\cos x - \sin x)}{(\sin x + \cos x)}$ cos2x=cos 2 x-sin 2 x

let sinx+cosx=t,(cosx-sinx)dx=dt

hence the given integral can be written as

$\int \frac{\cos 2 x}{(\sin x + \cos x)^{2}} d x = \int \frac{d t}{t} = l o g | t | + c = l o g | c o s x + s i n x | + c$

B is correct

Question 43: Choose the correct answers in Exercises 41 to 44.

If $f (a + b - x) = f (x)$ , then $\int_{a}^{b} x f (x) d x$ is equal to

(A) $\frac{a + b}{2} \int_{a}^{b} f (b - x) d x$

(B) $\frac{a + b}{2} \int_{a}^{b} f (b + x) d x$

(D) $\frac{a + b}{2} \int_{a}^{b} f (x) d x$

Answer:

$L e t \int_{a}^{b} x f (x) d x = I$

As we know $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

Using the above property we can write the integral as

$I = \int_{a}^{b} (a + b - x) f (a + b - x) d x I = \int_{a}^{b} (a + b - x) f (x) d x I = (a + b) \int_{a}^{b} f (x) d x - \int_{a}^{b} x f (x) d x I = (a + b) \int_{a}^{b} f (x) d x - I 2 I = (a + b) \int_{a}^{b} f (x) d x I = \frac{a + b}{2} \int_{a}^{b} f (x) d x$

Answer (D) is correct

Question 44: Choose the correct answers in Exercises 41 to 44.

The value of $\int_{0}^{1} \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$ is

(A) 1

(B) 0

(D) $\frac{π}{4}$

Answer:

$L e t I = \int_{0}^{1} \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x$

$t a n^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) = t a n^{- 1} (\frac{x - (1 - x)}{1 + x (1 - x)}) = t a n^{- 1} x - t a n^{- 1} (1 - x)$ as $t a n^{- 1} (\frac{a - b}{1 + a b}) = t a n^{- 1} a - t a n^{- 1} b$

Now the integral can be written as

$I = \int_{0}^{1} (t a n^{- 1} x - t a n^{- 1} (1 - x)) d x I = \int_{0}^{1} (t a n^{- 1} (1 - x) - t a n^{- 1} (1 - (1 - x))) d x I = \int_{0}^{1} (t a n^{- 1} (1 - x) - t a n^{- 1} x) d x I = - I 2 I = 0 I = 0$

(B) is correct.

Also, read

Topics covered in Chapter 7 Integrals: Miscellaneous Exercise

Some properties of indefinite integrals are as follows:

1. $\int [f (x) + g (x)] d x = \int f (x) d x + \int g (x) d x$

2. For any real number $k, \int k f (x) d x = k \int f (x) d x$

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C, n \neq - 1$ . Particularly, $\int d x = x + C$
$\int \cos x d x = \sin x + C$
$\int \sin x d x = - \cos x + C$
$\int \sec^{2} x d x = \tan x + C$
$\int {cosec}^{2} x d x = - \cot x + C$
$\int \sec x \tan x d x = \sec x + C$
$\int cosec x \cot x d x = - cosec x + C$
$\int \frac{d x}{\sqrt{1 - x^{2}}} = \sin^{- 1} x + C$
$\int \frac{d x}{\sqrt{1 - x^{2}}} = - \cos^{- 1} x + C$
$\int \frac{d x}{1 + x^{2}} = \tan^{- 1} x + C$
$\int \frac{d x}{1 + x^{2}} = - \cot^{- 1} x + C$
$\int e^{x} d x = e^{x} + C$
$\int a^{x} d x = \frac{a^{1}}{\log a} + C$
$\int \frac{1}{x} d x = \log | x | + C$
First fundamental theorem of integral calculus

Let the area function be defined by $A (x) = \int_{a}^{x} f (x) d x$ for all $x \geq a$ , where the function $f$ is assumed to be continuous on $[a, b]$ . Then $A^{'} (x) = f (x)$ for all $x \in [a, b]$ .

Second fundamental theorem of integral calculus

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Let $f$ be a continuous function of $x$ defined on the closed interval $[a, b]$ and let F be another function such that $\frac{d}{d x} F (x) = f (x)$ for all $x$ in the domain of $f$ , then $\int_{a}^{b} f (x) d x = [F (x) + C]_{a}^{b} = F (b) - F (a)$ .

This is called the definite integral of $f$ over the range $[a, b]$ , where $a$ and $b$ are called the limits of integration, $a$ being the lower limit and $b$ the upper limit.

Also, read,

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Students can refer to the NCERT Exemplar solutions from other subjects listed below for additional support and better concept clarity.

Frequently Asked Questions (FAQs)

1. What is the level of questions in Miscellaneous exercise ?

Advanced level problems are dealt with in this exercise.

2. How much time will it take topcomlete miscellaneous exercise for the first time ?

It can take 3 to 4 hours for the first time.

3. Can shortcuts be used while solving the questions.?

In boards, step by step method is used but shortcuts can be used in JEE and NEET.

4. Is it mandatory to solve every question of Miscellaneous exercise ?

No, but questions on different concepts must be solved.

5. What is the weightage from miscellaneous exercise in the examination ?

Around 5 marks of questions are asked in the examination.

6. Can this exercise be done with self study only ?

Yes, but for that basic concepts must be clear.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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Option 3) more than 6 but less than 9	Option 4) more than 9

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