RD Sharma Class 12 Exercise 18.22 Indefinite integrals Solutions Maths

RD Sharma Class 12 Exercise 18.22 Indefinite integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:26 PM IST

The Class 12 RD Sharma chapter 18 exercise 18.22 solution is the best solution for the chapter of Indefinite integrals for the students of class 12th. Students of any board can take these solutions for help if they want to score high in the exams. RD Sharma Solutions The RD Sharma class 12th exercise 18.22 is highly recommended for solving the complex chapters like Indefinite integrals, not only teachers but students as well prefer these solutions for thorough practice.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.22
Below are mentioned few benefits of the RD Sharma class 12 solution chapter 18 exercise 18.22 :-

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.22

Indefinite Integrals Exercise 18.22 Question 1.

Answer: $\frac{1}{6} \tan^{- 1} (\frac{3 \tan x}{2}) + c$
Given: $\int \frac{1}{4 \cos^{2} x + 9 \sin^{2} x} d x$
Hint: Use substitution method
Solution: $\int \frac{1}{4 \cos^{2} x + 9 \sin^{2} x} d x$
Dividing numerator and denominator by

\cos^{2} x

\begin{aligned} = \int \frac{\frac{1}{\cos^{2} x}}{\frac{4 \cos^{2} x}{\cos^{2} x} + \frac{9 \sin^{2} x}{\cos^{2} x}} d x \\ = \int \frac{\sec^{2} x d x}{4 + 9 \tan^{2} x} \end{aligned}

N o w, L e t

\begin{aligned} \tan x = t \\ \sec^{2} x d x = d t \end{aligned}

(Differentiating w.r.t to x)

So, \int \frac{d t}{4 + 9 t^{2}}

= \frac{1}{9} \int \frac{d t}{\frac{4}{9} + t^{2}}

= \frac{1}{9} \times \frac{3}{2} [\tan^{- 1} (\frac{t}{\frac{2}{3}})] + c

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

= \frac{1}{6} \tan^{- 1} (\frac{3 \tan x}{2}) + c

[\tan x = t]

Indefinite Integrals Exercise 18.22 Question 2.

Answer: $\frac{1}{2 \sqrt{5}} \tan^{- 1} (\frac{2 \tan x}{\sqrt{5}}) + c$
Given: $\int \frac{1}{4 \sin^{2} x + 5 \cos^{2} x} d x$
Hint: Divide the numerator and denominator by

\cos^{2} x

and then use substitution method
Solution:
$\int \frac{1}{4 \sin^{2} x + 5 \cos^{2} x} d x$
Dividing the numerator and denominator by

\cos^{2} x

\Rightarrow \int \frac{\frac{1}{\cos^{2} x}}{\frac{4 \sin^{2} x}{\cos^{2} x} + 5 \frac{\cos^{2} x}{\cos^{2} x}} d x

= \int \frac{\sec^{2} x}{4 \tan^{2} x + 5} d x

Let

\begin{matrix} t = \tan x \\ d t = \sec^{2} x d x \end{matrix}

(Differentiating w.r.t to x)
Now,

\int \frac{1}{4 t^{2} + 5} d t

= \frac{1}{4} \int \frac{1}{t^{2} + \frac{5}{4}} d t

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

= \frac{1}{4} \times \frac{2}{\sqrt{5}} \tan^{- 1} (\frac{t}{\frac{\sqrt{5}}{2}}) + c

= \frac{1}{2 \sqrt{5}} \tan^{- 1} (\frac{2 \tan x}{\sqrt{5}}) + c

Indefinite Integrals Exercise 18.22 Question 3.

Answer: $\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan x + 1}{\sqrt{3}}) + c$
Given: $\int \frac{2}{2 + \sin 2 x} d x$
Hint: You must know about substitution method
Solution: $\int \frac{2}{2 + \sin 2 x} d x$
$\begin{aligned} = \int \frac{2}{2 + 2 \sin x \cos x} d x \\ = \int \frac{1}{1 + \sin x \cos x} d x \end{aligned}$
Dividing the numerator and denominator by

\cos^{2} x

\begin{aligned} = \int \frac{\sec^{2} x}{\sec^{2} x + \tan x} d x \\ = \int \frac{\sec^{2} x}{1 + \tan^{2} x + \tan x} d x \end{aligned}

Let

\begin{aligned} \tan x = t \\ \sec^{2} d x = d t \end{aligned}

(Differentiating w.r.t x)
Now,

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

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

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

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

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

\begin{aligned} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{\sqrt{3}}) + c \\ = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan x + 1}{\sqrt{3}}) + c \end{aligned}

Indefinite Integrals Exercise 18.22 Question 4.

Answer: $\frac{1}{2 \sqrt{3}} \log | \frac{1 + \sqrt{3} \tan x}{1 - \sqrt{3} \tan x} | + c$
Given: $\int \frac{\cos x}{\cos 3 x} d x$
Hint: Use the formula of

\cos 3 x

Solution:
$\int \frac{\cos x}{\cos 3 x} d x$
$= \int \frac{\cos x}{4 \cos^{3} x - 3 \cos x} d x$
$(∵ \cos 3 x = 4 \cos^{3} x - 3 \cos x)$
$= \int \frac{1}{4 \cos^{2} x - 3} d x$
Dividing numerator and denominator by

\cos^{2} x

\begin{aligned} = \int \frac{\sec^{2} x}{4 - 3 \sec^{2} x} d x \\ = \int \frac{\sec^{2} x}{4 - 3 (1 + \tan^{2} x)} d x \end{aligned}

(∵ \sec^{2} x = 1 + \tan^{2} x)

= \int \frac{\sec^{2} x}{1 - 3 \tan^{2} x} d x

Let

\begin{aligned} \tan x = t \\ \sec^{2} x d x = d t \end{aligned}

(Differentiating w.r.t x)

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

= \frac{1}{3} \times \frac{1}{2} \times \frac{1}{\frac{1}{\sqrt{3}}} \log | \frac{\frac{1}{\sqrt{3}} + t}{\frac{1}{\sqrt{3}} - t} | + c

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

= \frac{1}{6} \times \sqrt{3} \log | \frac{1 + \sqrt{3} t}{1 - \sqrt{3} t} | + c

= \frac{1}{2 \sqrt{3}} \log | \frac{1 + \sqrt{3} \tan x}{1 - \sqrt{3} \tan x} | + c

Indefinite Integrals Exercise 18.22 Question 5.

Answer: $\frac{1}{2} \tan^{- 1} (2 \tan x) + c$
Given: $\int \frac{1}{1 + 3 \sin^{2} x} d x$
Hint: Use the formula of

\sec^{2} x

and then apply substitution method
Solution: $\int \frac{1}{1 + 3 \sin^{2} x} d x$
Divide numerator and denominator by

\cos^{2} x

= \int \frac{\frac{1}{\cos^{2} x}}{\frac{1}{\cos^{2} x} + \frac{3 \sin^{2} x}{\cos^{2} x}} d x

= \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x} d x

= \int \frac{\sec^{2} x}{1 + \tan^{2} x + 3 \tan^{2} x} d x

(\sec^{2} x = 1 + \tan^{2} x)

= \int \frac{\sec^{2} x}{1 + 4 \tan^{2} x} d x

L e t

t = \tan x

(Differentiating w.r.t x)

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

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

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

= \frac{1}{4} \times \frac{1}{\frac{1}{2}} \tan^{- 1} (\frac{t}{\frac{1}{2}}) + c

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

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

Indefinite Integrals Exercise 18.22 Question 6.

Answer: $\frac{1}{\sqrt{15}} \tan^{- 1} (\frac{\sqrt{3} \tan x}{\sqrt{5}}) + c$
Given: $\int \frac{1}{3 + 2 \cos^{2} x} d x$
Hint: Divide Numerator and Denominator by

\cos^{2} x

and then apply substitution method.
Solution: $\int \frac{1}{3 + 2 \cos^{2} x} d x$
On dividing Numerator and Denominator by

\cos^{2} x

= \int \frac{\sec^{2} x}{3 \sec^{2} x + 2}

= \int \frac{\sec^{2} x}{3 (1 + \tan^{2} x) + 2} d x

(\sec^{2} x = 1 + \tan^{2} x)

= \int \frac{\sec^{2} x}{5 + 3 \tan^{2} x} d x

L e t

t = \tan x

(Differentiating w.r.t x)

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

N o w

\int \frac{1}{5 + 3 t^{2}} d t

= \frac{1}{3} \int \frac{1}{({(\sqrt{\frac{5}{3}})}^{2} + t^{2})} d t

= \frac{1}{3} \times \frac{\sqrt{3}}{\sqrt{5}} \tan^{- 1} (\frac{t}{\sqrt{\frac{5}{3}}}) + c

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

= \frac{1}{\sqrt{15}} \tan^{- 1} (\frac{\sqrt{3} \tan x}{\sqrt{5}}) + c

Indefinite Integrals Exercise 18.22 Question 7.

Answer: $\frac{1}{5} \log | \frac{\tan x - 2}{2 \tan x + 1} | + c$
Given: $\int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x$
Hint: Apply substitution method and divide Numerator and Denominator by

\cos^{2} x

Solution:
$\int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x$
$= \int \frac{1}{2 \sin^{2} x + \sin x \cos x - 4 \sin x \cos x - 2 \cos^{2} x} d x$
$= \int \frac{1}{2 \sin^{2} x - 3 \sin x \cos x - 2 \cos^{2} x} d x$
On dividing Numerator and Denominator by

\cos^{2} x

= \int \frac{\sec^{2} x}{2 \tan^{2} x - 3 \tan x - 2} d x

L e t

\tan x = t

(Differentiate w.r.t x)

\sec^{2} d x = d t

N o w,

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

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

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

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

= \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \log | \frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}} | + c

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

\begin{aligned} = \frac{1}{2} \times \frac{2}{5} \log | \frac{2 t - 4}{2 t + 1} | + c \\ = \frac{1}{5} \log | \frac{2 \tan x - 4}{2 \tan x + 1} | + c \end{aligned}

Indefinite Integrals Exercise 18.22 Question 8.

Answer: $\tan^{- 1} (\tan^{2} x) + c$
Given: $\int \frac{\sin 2 x}{\sin^{4} x + \cos^{4} x} d x$
Hint: Divide Numerator and denominator by

\cos^{4} x

Solution:
$\int \frac{\sin 2 x}{\sin^{4} x + \cos^{4} x} d x$ $(\sin 2 x = 2 \sin x \cos x)$
$= \int \frac{2 \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$
On dividing Numerator and denominator by

\cos^{4} x

= \int \frac{2 \tan x \cdot \sec^{2} x d x}{\tan^{4} x + 1}

L e t

\begin{aligned} \tan^{2} x = t \\ 2 \tan x \sec^{2} x d x = d t \end{aligned}

(Differentiate w.r.t x)

N o w,

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

= \tan^{- 1} t + c

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

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

Indefinite Integrals Exercise 18.22 Question 10.

Answer: $\frac{1}{2} \log | \frac{\tan x}{\tan x + 2} | + c$
Given: $\int \frac{1}{\sin^{2} x + \sin 2 x} d x$
Hint: Use the formula of sin 2x and then apply substitution method
Solution: $\int \frac{1}{\sin^{2} x + \sin 2 x} d x$
$= \int \frac{1}{\sin^{2} x + 2 \sin x \cos x} d x$ $(\sin 2 x = 2 \sin x \cos x)$
On dividing numerator and denominator by

\cos^{2} x

, we get

= \int \frac{\sec^{2} x}{\tan^{2} x + 2 \tan x} d x

L e t

t = \tan x

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

(Differentiate w.r.t x)

N o w,

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

\begin{aligned} = \int \frac{1}{t^{2} + 2 t + (1)^{2} - (1)^{2}} d t \\ = \int \frac{1}{(t + 1)^{2} - (1)^{2}} d t \end{aligned}

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

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

\begin{aligned} = \frac{1}{2} \log | \frac{t}{t + 2} | + c \\ = \frac{1}{2} \log | \frac{\tan x}{\tan x + 2} | + c \end{aligned}

Indefinite Integrals Exercise 18.22 Question 11.

Answer: $\frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{2} \tan x) + c$
Given: $\int \frac{1}{\cos 2 x + 3 \sin^{2} x} d x$
Hint: Divide numerator and denominator by

\cos^{2} x

and then use substitution method
Solution:
$\begin{aligned} \int \frac{1}{\cos 2 x + 3 \sin^{2} x} d x \\ (\cos 2 x = \cos^{2} x - \sin^{2} x) \\ = \int \frac{1}{\cos^{2} x - \sin^{2} x + 3 \sin^{2} x} d x \\ = \int \frac{1}{\cos^{2} x + 2 \sin^{2} x} d x \end{aligned}$
On dividing numerator and denominator by

\cos^{2} x

, we get

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

L e t

\begin{aligned} \tan x = t \\ \sec^{2} x d x = d t \end{aligned}

(Differentiate w.r.t x)

N o w,

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

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

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

= \frac{1}{2} \times \frac{\sqrt{2}}{1} \tan^{- 1} (\frac{t}{\frac{1}{\sqrt{2}}}) + c

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

RD Sharma class 12 solution of Indefinite Integrals exercise 18.22 contains 11 questions, covered the following topics that are mentioned below:-

Integration of trigonometric and inverse trigonometric functions
Integration of exponential functions

Below are mentioned few benefits of the RD Sharma class 12 solution chapter 18 exercise 18.22 :-

The RD Sharma class 12th exercise 18.22 solution is crafted by great experts that are famous in the field of mathematics and thus it gives a vast proportion of knowledge to students in a way that no school can provide.
The syllabus of NCERT is quite vast and difficult, therefore students face trouble while solving questions from it and that is why RD Sharma class 12 chapter 18 exercise 18.22 is recommended for solved questions provided in it.
The most important benefit of downloading the RD Sharma class 12th exercise 18.22 from the Career360 website is that it is available free of cost for download on the website.
The RD Sharma solutions are used by teachers for lectures and also to provide homework so that makes it easy for students if they practice from the solution they can always stay ahead of the class.
RD Sharma class 12th exercise 18.22 gives a brief idea about the concepts and explains it in easy, moderate and tough level which will prepare you to solve any type of question asked in the exam.

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Frequently Asked Questions (FAQs)

1. What are indefinite integrals?

It is an integral expressed without limits, and so containing an arbitrary constant.

2. What is the general formula for indefinite integrals?

Either way, the general antiderivative is x+c and so 1dx = x+c, where c is a constant.

3. What is the process of finding an integral called?

The process of finding an integral is also called integration.

4. Is the RD Sharma solution available online?

Yes, you can download the RD Sharma solution from the official website of Careers360

5. How many questions are there in RD Sharma exercise 18.28?

There are a total of 11 questions in RD Sharma exercise 18.28 for helpful knowledge

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RD Sharma Class 12 Exercise 18.22 Indefinite integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.22

Below are mentioned few benefits of the RD Sharma class 12 solution chapter 18 exercise 18.22 :-

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