RD Sharma Class 12 Exercise 18.18 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.18 Indefinite Integrals Solutions Maths - Download PDF Free Online

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The Rd Sharma Class 12th Exercise 18.18 solutions are the best to buy if a student is preparing for competitive exams like IIT JEE. The chapters and exercises in RD Sharma Class 12 are formatted in such a way that a student may find it difficult to understand. The students will be able to grasp the concepts with RD Sharma Class 12 Solutions chapter 18 Exercise 18.18. Rd Sharma Class 12th Exercise 18.18. RD Sharma Solutions All the questions and answers from the chapter of Indefinite Integrals are designed to help students learn the concepts clearly to achieve a higher score in any exam.

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This exercise has all the questions related to integrating the following function by single integration, double integral or multiple integrals. This specific exercise has fourteen questions in level one and four questions in level two.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.18

Indefinite Integrals Exercise 18.18 Question 1

Answer: $\frac{1}{2} \log | x^{2} + \sqrt{x^{4} + a^{4}} | + c$
Hint: $\int \sqrt{x^{2} + a^{2}} = \log | x + \sqrt{x^{2} + a^{2}} | + c$
Given: $\int \frac{x}{\sqrt{x^{4} + a^{4}}} d x$ .............(1)
Explanation:

$L e t x^{2} = t$
$2 x d x = d t$
$x d x = \frac{d t}{2}$ (Differentiate w.r.t to t)
Put in (1)

\frac{1}{2} \int \frac{d t}{\sqrt{t^{2} + {(a^{2})}^{2}}}

[∵ \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \log | x + \sqrt{x^{2} + a^{2}} | + c]

= \frac{1}{2} \log | t + \sqrt{t^{2} + a^{4}} | + c

= \frac{1}{2} \log | x^{2} + \sqrt{x^{4} + a^{4}} | + c

Indefinite Integrals Exercise 18 Point 18 Question 2

Answer: $\log | \tan x + \sqrt{4 + \tan^{2} x} | + c$
Hint: Let

\tan x = t

Given: $\int \frac{\sec^{2} x d x}{\sqrt{4 + \tan^{2} x}}$
Explanation: $\int \frac{\sec^{2} x d x}{\sqrt{4 + \tan^{2} x}}$
Let

\tan x = t

\sec^{2} x d x = d t

(Differentiate w.r.t to t)

= \int \frac{d t}{\sqrt{4 + t^{2}}}

= \int \frac{d t}{\sqrt{(2)^{2} + t^{2}}}

= \log | t + \sqrt{4 + t^{2}} | + c

[∵ \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \log | x + \sqrt{x^{2} + a^{2}} | + c]

= \log | \tan x + \sqrt{4 + \tan^{2} x} | + c

Indefinite Integrals Exercise 18 Point 18 Question 3

Answer: $\sin^{- 1} (\frac{e^{x}}{4}) + c$
Hint $L e t e^{x} = t$
Given: $\int \frac{e^{x}}{\sqrt{16 - e^{2 x}}} d x$
Explanation:
$\int \frac{e^{x}}{\sqrt{16 - e^{2 x}}} d x$ ..........(1)
$L e t e^{x} = t$
$e^{x} d x = d t$
From (1) we have

\int \frac{d t}{\sqrt{(4)^{2} - t^{2}}}

[∵ \int \frac{d t}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a}]

= \sin^{- 1} (\frac{t}{4}) + c

= \sin^{- 1} (\frac{e^{x}}{4}) + c

Indefinite Integrals Exercise 18.18 Question 4

Answer:

\log | \sin x + \sqrt{4 + \sin^{2} x} | + c

Hint:Let

\sin x = t

Given: $\int \frac{\cos x}{\sqrt{4 + \sin^{2} x}} d x$
Explanation:
$\int \frac{\cos x}{\sqrt{4 + \sin^{2} x}} d x$ .......(1)
$l e t \sin x = t$ (Differentiate w.r.t to t)

\cos d x = d t

From (1) we have

\int \frac{d t}{\sqrt{(2)^{2} + t^{2}}} = \log | t + \sqrt{4 + t^{2}} | + c [∵ \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \log | x + \sqrt{x^{2} + a^{2}} | + c]

= \log | \sin x + \sqrt{4 + \sin^{2} x} | + c

Indefinite Integrals Exercise 18.18 Question 5

Answer: $\frac{- 1}{2} \log | 2 \cos x + \sqrt{4 \cos^{2} x - 1} | + c$
Hint Let

2 \cos x = t

Given: $\int \frac{\sin x}{\sqrt{4 \cos^{2} x - 1}} d x$
Explanation:
$\int \frac{\sin x}{\sqrt{4 \cos^{2} x - 1}} d x$ ..........(1)
$L e t 2 \cos x = t$
$- 2 \sin x d x = d t$
$\sin x d x = \frac{- 1}{2} d t$ (Differentiate w.r.t to t)
Put in (1) we get

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

= \frac{- 1}{2} \log | t + \sqrt{t^{2} - 1} + c | [∵ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]

= \frac{- 1}{2} \log | 2 \cos x + \sqrt{4 \cos^{2} x - 1} | + c

Indefinite Integrals Exercise 18.18 Question 7

Answer:

\frac{1}{3} \sin^{- 1} (\frac{3 \log x}{2}) + c

Hint: Let

3 \log x = 2

Given: $\int \frac{1}{x \sqrt{4 - 9 (\log x)^{2}}} d x$
Explanation:
$\int \frac{1}{x \sqrt{4 - 9 (\log x)^{2}}} d x$ ........(1)
Let

3 \log x = t

\frac{3}{x} d x = d t \Rightarrow \frac{d x}{x} = \frac{d t}{3}

Put in (1)

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

= \frac{1}{3} \sin^{- 1} (\frac{t}{2}) + c [\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \sin^{- 1} (\frac{x}{a}) + c]

= \frac{1}{3} \sin^{- 1} (\frac{3 \log x}{2}) + c

Indefinite Integrals Exercise 18.18 Question 8

Answer: $\frac{1}{4} \log | \sin^{2} 4 x + \sqrt{9 + \sin^{4} 4 x} | + c$
Hint Let

\sin^{2} 4 x = t

Given: $\int \frac{\sin 8 x}{\sqrt{9 + \sin^{4} 4 x}} d x$
Put
$\sin^{2} 4 x = t \Rightarrow 2 \times 4 \sin 4 x \cdot \cos 4 x d x = d t$
$\sin 8 x d x = \frac{1}{4} d t$

\Rightarrow I = \frac{1}{4} \int \frac{1}{\sqrt{9 + t^{2}}} d t = \log | t + \sqrt{9 + t^{2}} + | + c \Rightarrow I = \frac{1}{4} \log | \sin^{2} 4 x + \sqrt{9 + \sin^{4} 4 x} | + c

Indefinite Integrals Exercise 18.18 Question 9

Answer: $\frac{1}{2} \log | \sin 2 x + \sqrt{\sin^{2} 2 x + 8} | + c$
Hint

\sin 2 x = t

Given: $\int \frac{\cos 2 x}{\sqrt{\sin^{2} 2 x + 8}} d x$
Explanation:
$\int \frac{\cos 2 x}{\sqrt{\sin^{2} 2 x + 8}} d x$ ...........(1)
Let

\sin 2 x = t

2 \cos 2 x d x = d t

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

(Differentiate w.r.t to t)
Put in (1), we have

\frac{1}{2} \int \frac{d t}{\sqrt{t^{2} + 8}}

= \frac{1}{2} \log | t + \sqrt{t^{2} + 8} | + c [∵ \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \log | x + \sqrt{x^{2} + a^{2}} | + c]

= \frac{1}{2} \log | \sin 2 x + \sqrt{\sin^{2} 2 x + 8} | + c

Indefinite Integrals Exercise 18.18 Question 10

Answer: $\log | \sin^{2} x + 2 + \sqrt{\sin^{4} x + 4 \sin^{2} x - 2} | + c$
Hint Let

\sin^{2} x = t

Given: $\int \frac{\sin 2 x}{\sqrt{\sin^{4} x + 4 \sin^{2} x - 2}} d x$
Explanation:
$\int \frac{\sin 2 x}{\sqrt{\sin^{4} x + 4 \sin^{2} x - 2}} d x$ .................(1)
Let

\sin^{2} x = t

2 \sin x \cos x d x = d t

\sin 2 x d x = d t

(Differentiate w.r.t to t)
From (1) we have

\int \frac{d t}{\sqrt{t^{2} + 4 t - 2}}

= \int \frac{d t}{\sqrt{t^{2} + 4 t + 4 - 4 - 2}}

= \int \frac{d t}{\sqrt{(t + 2)^{2} - 6}}

Let t+2 = u (Differentiate w.r.t to u)

dt = du

= \int \frac{d u}{\sqrt{u^{2} - 6}}

= \log | u + \sqrt{u^{2} - 6} | + c

= \log | t + 2 + \sqrt{(t + 2)^{2} - 6} | + c

= \log | \sin^{2} x + 2 + \sqrt{\sin^{4} x + 4 \sin^{2} x - 2} | + c [∵ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]

Indefinite Integrals Exercise 18.18 Question 11

Answer: $- \log | (\cos^{2} x + \frac{1}{2}) + \sqrt{\cos^{4} x + \cos^{2} x + 1} | + c$
Hint Let

\cos^{2} x = t

Given: $\int \frac{\sin 2 x}{\sqrt{\cos^{4} x - \sin^{2} x + 2}} d x$
Explanation:
$\int \frac{\sin 2 x}{\sqrt{\cos^{4} x - \sin^{2} x + 2}} d x$
Let

\cos^{2} x = t

- 2 \cos x \sin x d x = d t

- \sin 2 x d x = d t

(Differentiate w.r.t to t)
From (1) we have

- \int \frac{d t}{\sqrt{t^{2} - (1 - t) + 2}}

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

= - \int \frac{d t}{\sqrt{t^{2} + 2 \cdot \frac{1}{2} \cdot t + \frac{1}{4} - \frac{1}{4} + 1}}

= \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} + \frac{3}{4}}}

= - \log | t + \frac{1}{2} + \sqrt{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} | + c

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

Indefinite Integrals Exercise 18.18 Question 12

Answer: $\sin^{- 1} (\frac{\sin x}{2}) + c$
Hint $\sin x = t$
Given: $\int \frac{\cos x}{\sqrt{4 - \sin^{2} x}} d x$
Explanation:
$\int \frac{\cos x}{\sqrt{4 - \sin^{2} x}} d x$ ..............(1)
Let

\sin x = t

\cos x d x = d t

(Differentiate w.r.t to t)
From (1) we have

\int \frac{d t}{\sqrt{4 - t^{2}}}

= \sin^{- 1} (\frac{t}{2}) + c [∵ \int \frac{d t}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a}]

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

Indefinite Integrals Exercise 18.18 Question 13

Answer: $3 \log | x^{\frac{1}{3}} + \sqrt{x^{\frac{2}{3}} - 4} | + c$
Hint Let

x^{\frac{1}{3}} = t

Given: $\int \frac{1}{x^{\frac{2}{3}} \sqrt{x^{\frac{2}{3}} - 4}} d x$
Explanation:
$\int \frac{1}{x^{\frac{2}{3}} \sqrt{x^{\frac{2}{3}} - 4}} d x$ ............(1)
Let

x^{\frac{1}{3}} = t

\frac{1}{3} x^{\frac{- 2}{3}} d x = d t

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

(Differentiate w.r.t to t)
Put in (1) we have

3 \int \frac{d t}{\sqrt{t^{2} - 4}}

= 3 \log | t + \sqrt{t^{2} - 4} | + c

= 3 \log | x^{\frac{1}{3}} + \sqrt{x^{\frac{2}{3}} - 4} | + c [∵ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]

Indefinite Integrals Exercise 18.18 Question 14

Answer: $\log | \sin^{- 1} x + \sqrt{9 + {(\sin^{- 1} x)}^{2}} | + c$
Hint Let

\sin^{- 1} x = t

Given: $\int \frac{1}{\sqrt{(1 - x^{2}) (9 + {(\sin^{- 1} x)}^{2})}} d x$
Explanation:
$\int \frac{1}{\sqrt{(1 - x^{2}) (9 + {(\sin^{- 1} x)}^{2})}} d x$ .........(1)
Let

\sin^{- 1} x = t

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

From (1) we have

\int \frac{d t}{\sqrt{9 + t^{2}}}

= \log | t + \sqrt{9 + t^{2}} | + c

= \log | \sin^{- 1} x + \sqrt{9 + {(\sin^{- 1} x)}^{2}} | + c

From (1) we have

Indefinite Integrals Exercise 18.18 Question 15

Answer: $\log | (\sin x - 1) + \sqrt{\sin^{2} x - 2 \sin x - 3} | + c$
Hint Let

\sin x = t

Given: $\int \frac{\cos x}{\sqrt{\sin^{2} x - 2 \sin x - 3}} d x$
Explanation:
$\int \frac{\cos x}{\sqrt{\sin^{2} x - 2 \sin x - 3}} d x$ ...........(1)
Let

\sin x = t

Cos x d x = d t

Put in (1) we have

\int \frac{d t}{\sqrt{t^{2} - 2 t - 3}}

= \int \frac{d t}{\sqrt{t^{2} - 2 t + 1 - 1 - 3}}

= \int \frac{d t}{\sqrt{t^{2} - 2 t + 1 - 4}}

= \int \frac{d t}{\sqrt{(t - 1)^{2} - (2)^{2}}}

= \log | t - 1 + \sqrt{(t - 1)^{2} - (2)^{2}} | + c [∵ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]

= \log | (\sin x - 1) + \sqrt{\sin^{2} x - 2 \sin x - 3} | + c

Indefinite Integrals Exercise 18.18 Question 16

Answer: $\log (\sin x + \frac{1}{2}) + \sqrt{\sin^{2} x + \sin x} ∣ + c$
Hint Let

\sin x = t

Given: $\int \sqrt{cosec x - 1} d x$
Explanation:
$\int \sqrt{\cos e c x - 1} d x$
$= \int \sqrt{\frac{1}{\sin x} - 1} d x$
$= \int \sqrt{\frac{1 - \sin x}{\sin x}} d x$
Multiply and Divide with (1 + sin x)

= \int \sqrt{\frac{(1 - \sin x) (1 + \sin x)}{\sin x (1 + \sin x)}} d x

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

[\begin{array}{l} \sin^{2} x + \cos^{2} x = 1 \\ \cos^{2} x = 1 - \sin^{2} x \end{array}]

= \int \sqrt{\frac{\cos^{2} x}{\sin x (1 + \sin x)}} d x

= \int \frac{\cos x}{\sqrt{\sin x (1 + \sin x)}} d x

...........(1)
Let

\sin x = t

Cos x d x = d t

From (1)

\int \frac{d t}{\sqrt{t (1 + t)}}

= \int \frac{d t}{\sqrt{t^{2} + t + \frac{1}{4} - \frac{1}{4}}}

= \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}}

= \log (t + \frac{1}{2}) + \sqrt{t^{2} + t} ∣ + c [∵ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]

= \log | (\sin x + \frac{1}{2}) + \sqrt{\sin^{2} x + \sin x} | + c

Indefinite Integrals Exercise 18.18 Question 17

Answer: $- \log | (\sin x + \cos x) + \sqrt{\sin 2 x} | + c$
Hint $\sin^{2} x + \cos^{2} x = 1$
Given: $\int \frac{\sin x - \cos x}{\sqrt{\sin 2 x}} d x$
Explanation:
$\int \frac{\sin x - \cos x}{\sqrt{1 + \sin 2 x - 1}} d x$
$\int \frac{\sin x - \cos x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x - 1}} d x$ $[\begin{array}{l} \sin^{2} x + \cos^{2} x = 1 \\ \sin 2 x = 2 \sin x \cdot \cos x \end{array}]$
$\int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^{2} - 1}} d x$ ...........(1) $[(a + b)^{2} = a^{2} + b^{2} + 2 a b]$
Let

\sin x + \cos x = t

(\cos x - \sin x) d x = d t

- (\sin x - \cos x) d x = d t

From (1) we have

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

= - \log | t + \sqrt{t^{2} - 1} | + c

[∵ \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \log | x + \sqrt{x^{2} - a^{2}} | + c]

= - \log | \sin x + \cos x + \sqrt{(\sin x + \cos x)^{2} - 1} | + c

= - \log | \sin x + \cos x + \sqrt{\sin 2 x} | + c

Indefinite Integrals Exercise 18.18 Question 18

Answer:

\sin^{- 1} [\frac{1}{3} (\sin x + \cos x)] + c

Hint

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

Given: $\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2 x}} d x$
Explanation:
$\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2 x}} d x$
$= \int \frac{\cos x - \sin x}{\sqrt{8 + 1 - 1 - \sin 2 x}} d x$
$= \int \frac{\cos x - \sin x}{\sqrt{9 - (\sin^{2} x + \cos^{2} x + 2 \sin x \cos x)}} d x [\begin{array}{l} \sin^{2} x + \cos^{2} x = 1 \\ \sin 2 x = 2 \sin x \cdot \cos x \end{array}]$
$= \int \frac{\cos x - \sin x}{\sqrt{9 - (\sin x + \cos x)^{2}}} d x$
Let

\cos x + \sin x = t

(- \sin x + \cos x) d x = d t

Put in (1)

\int \frac{d t}{\sqrt{3^{2} - t^{2}}} = \sin^{- 1} [\frac{t}{3}] + c

[\int \frac{d t}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a}]

= \sin^{- 1} [\frac{1}{3} (\sin x + \cos x)] + c

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