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

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:19 PM IST

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

Rd Sharma Class 12th Exercise 18.18 solutions are here to help students to understand, learn and hold a good depth on the subject. The uses of these solutions are:

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Rd Sharma Class 12th Exercise 18.18 solutions are created by the group of subject experts. The team takes ample time to solve a question and brings out different ways to solve them. Therefore, a student will always be able to solve every question in mathematics after going for Rd Sharma Class 12th Chapter 18 Exercise 18.18 solutions.

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Rd Sharma Class 12th Exercise 18.18 solutions help students to prepare in a proper strategic manner for exams as every concept is understood step by step that makes a student write the exams confidently. The students will be able to understand exam patterns with RD Sharma Class 12th Solutions Indefinite Integrals Ex. 18.18.

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RD Sharma Class 12 Solutions Indefinite Integrals Ex. 18.18 solutions help a student do the homework and as these solutions are based on NCERT questions that are teacher’s favorite. RD Sharma Class 12 Solutions Chapter 18 ex 18.18 solutions help students achieve heights in career.

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RD Sharma Class 12th Solutions Chapter 18 ex 18.18 helps students find different ways to solve a problem. The concepts are connected with each other and solved by experts so that a student understands and solves questions in a different manner.

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Class 12th RD Sharma Chapter 18 Exercise 18.18 Solutions help a student learn the sums and understand ways to solve them. RD Sharma Class 12 Solutions Ex. 18.18 covers all the important solutions that are likely to appear in board and school exams.

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RD Sharma Chapter-wise Solutions

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1. Which of the following is an RD Sharma Chapter 18.18 exercise for Class 12?

The RD Sharma Solutions Class 12 Chapter 18 Exercise 18.18 Indefinite Integrals will assist students in developing a better understanding of the types of questions that are commonly asked in exams from this topic. Students should download and review Class 12 Maths Chapter 18 Exercise 18.18 to gain a thorough understanding of the concepts of Indefinite Integrals.

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

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.18

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