RD Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	19
Exercise	19.22
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals

Question 1. Evaluate the integral:

$\int\frac{1}{4cos^2x+9sin^2x}dx$

Solution:

Let $I=\int\frac{1}{4cos^2x+9sin^2x}dx$

On dividing numerator and denominator by cos²x, we get

$=\int\frac{\frac{1}{cos^2x}}{4+9tan^2x}dx\\ I=\int\frac{sec^2x}{4+9tan^2x}dx\\$

Let us considered tan x = t

So, sec²x dx = dt

$I=\int\frac{dt}{4+9(t)^2}\\ =\int\frac{dt}{4+(3t)^2}$

Again, let us considered 3t = u

3dt = du

$I=\frac{1}{3}\int\frac{du}{(2)^2+(u)^2}$

= (3/2) × (1/2) × tan^-1(u/2) + c

= (1/6)tan^-1(3t/2) + c

Hence, I = (1/6)tan^-1(3tanx/2) + c

Question 2. Evaluate the integral:

$\int\frac{1}{4sin^2x+5cos^2x}dx$

Solution:

Let $I=\int\frac{1}{4sin^2x+5cos^2x}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{\frac{1}{cos^2x}}{4tan^2x+5}dx\\ =\int\frac{sec^2x}{4tan^2x+5}dx$

Now, let us considered tan x = t

So, sec²xdx = dt

$I=\int\frac{dt}{4+9(t)^2}\\ =\int\frac{dt}{4t^2+5}$

Again, let us considered 2t = u

2dt = du

$I=\frac{1}{2}\int\frac{du}{(4)^2+(\sqrt{5})^2}$

= (1/2) × (1/√5) × tan^-1(u/√5) + c

= (1/2√5) × tan^-1(2t/√5) + c

Hence, I = (1/2√5) × tan^-1(2tanx/√5) + c

Question 3. Evaluate the integral:

$\int\frac{2}{2+sin2x}dx$

Solution:

Let $I=\int\frac{2}{2+sin2x}dx\\ =\int\frac{2}{2+2sinx\ cosx}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{\frac{1}{cos^2x}}{\frac{1}{cos^2x}+\frac{sinx\ cosx}{cos^2x}}dx\\ =\int\frac{sec^2x}{sec^2x+tanx}dx\\ I=\int\frac{sec^2x}{1+tan^2x+tanx}dx$

Now, let us considered tan x = t

So, sec²x dx = dt

$I=\int\frac{dt}{t^2+t+1}\\ =\int\frac{dt}{t^2+2t\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1}\\ I=\int\frac{dt}{\left(t+\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}\right)^2}\\ =\frac{1}{\frac{\sqrt3}{2}}tan^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt3}{2}}\right)+c\\ =\frac{2}{\sqrt3}tan^{-1}\left(\frac{2t+1}{\sqrt3}\right)+c\\ I=\frac{2}{\sqrt{3}}tan^{-1}\left(\frac{2tanx+1}{\sqrt3}\right)+c$

Question 4. Evaluate the integral:

$\int\frac{cosx}{cos3x}dx$

Solution:

Let $I=\int\frac{cosx}{cox3x}dx\\ =\int\frac{cosx}{4cos^3x-3cosx}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{\frac{cosx}{cos^3x}}{\frac{4cos^3x}{cos^3x}+\frac{3cosx}{cos^3x}}dx\\ =\int\frac{sec^2x}{4-3sec^2x}dx\\ =\int\frac{sec^2x}{4-3(1+tan^2x)}dx\\ =\int\frac{sec^2x}{4-3-3tan^2x}dx\\ =\int\frac{sec^2x}{1-3tan^2x}dx$

Now, let us considered tan x = t

So, sec²x dx = dt

$I=\int\frac{dt}{1-3t^2}\\ =\int\frac{dt}{1-(\sqrt3t)^2}$

Again, let us considered √3t = u

So, √3dt = du

$=\int\frac{du}{(1)^2-(4)^2}\\ =\frac{1}{2\sqrt3}\log\left|\frac{u+1}{1-u}\right|+c\\ =\frac{1}{2\sqrt3}\log\left|\frac{\sqrt3+1}{1-\sqrt3t}\right|+c\\ I=\frac{1}{2\sqrt3}\log\left|\frac{1+\sqrt3tanx}{1-\sqrt3tanx}\right|+c$

Question 5. Evaluate the integral:

$\int\frac{1}{1+3sin^2x}dx$

Solution:

Let $I=\int\frac{1}{1+3sin^2x}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{\frac{1}{cos^2x}}{\frac{1}{cos^2x}+\frac{3sin^2x}{cos^2x}}dx\\ =\int\frac{sec^2x}{sec^2x+3tan^2x}dx\\ =\int\frac{sec^2x}{1+tan^2x+3tan^2x}dx\\ =\int\frac{sec^2x}{1+\{2tanx\}^2}dx\\ =\int\frac{sec^x}{1+\{2tanx\}^2}dx$

Now, let us assume 2tan x = t

So, 2sec²x dx = dt

I = 1/2 ∫dt/(1 + t²)

= 1/2 tan^-1t + c

Hence, I = 1/2 tan^-1(2tanx) + c

Question 6. Evaluate the integral:

$\int\frac{1}{3+2cos^2x}dx$

Solution:

Let $I=\int\frac{1}{3+2cos^2x}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{\frac{1}{cos^2x}}{\frac{3}{cos^2x}+\frac{2cos^2x}{cos^2x}}dx\\ =\int\frac{sec^2x}{3sec^2x+2}dx\\ =\int\frac{sec^2x}{3(1+tan^2x)+2}dx\\ =\int\frac{sec^2x}{3+3tan^2x+2}dx\\ =\int\frac{sec^2x}{5+3tan^2x+2}dx$

Now, let us assume √3 tanx = t

So, √3 sec²x dx = dt

$I=\frac{1}{\sqrt3}\int\frac{dt}{(\sqrt5)^2+t^2}\\ =\frac{1}{\sqrt3\times\sqrt5}tan^{-1}\left(\frac{t}{\sqrt5}\right)+c\\$

Hence, I = (1/√15)tan^-1(√3tanx/√5) + c

Question 7. Evaluate the integral:

$\int\frac{1}{(sinx-2cosx)(2sinx+cosx)}dx$

Solution:

Let $I=\int\frac{1}{(sinx-2cosx)(2sinx+cosx)}dx\\ =\int\frac{1}{2sin^2x+sinxcosx-4sinxcosx-2cos^2x}dx\\ =\int\frac{1}{2sin^2x-3sinxcosx-2cos^2x}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{sec^2x}{2tan^2x-3tanx-2}dx$

Now, let us assume tanx = t

So, sec²x dx = dt

$I=\int\frac{dt}{2t^2-3t-2}\\ =\frac{1}{2}\int\frac{dt}{t^2-\frac{3}{2}t-1}\\ =\frac{1}{2}\int\frac{dt}{t^2-2t\left(\frac{3}{4}\right)+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2-1}\\ I=\frac{1}{2}\int\frac{dt}{\left(t-\frac{3}{4}\right)^2-\left(\frac{5}{4}\right)^2}\\ =\frac{1}{2}\times\frac{1}{2\frac{5}{4}}\log\left|\frac{t-\frac{3}{4}-\frac{5}{4}}{t-\frac{3}{4}+\frac{5}{4}}\right|+c\\ =\frac{1}{5}\log\left|\frac{t-2}{2t+1}\right|+c\\ I=\frac{1}{5}\log\left|\frac{tanx-2}{2tanx+1}\right|+c$

Question 8. Evaluate the integral:

$\int\frac{sin2x}{sin^4x+cos^4x}dx$

Solution:

Let $I=\int\frac{sin2x}{sin^4x+cos^4x}dx$

On dividing numerator and denominator by cos⁴x, we get

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

Now, let us assume tan²x = t

So, 2tanx sec²x dx = dt

I = ∫dt/(t²+ 1)

= tan^-1t + c

I = tan^-1(tan²x) + c

Question 9. Evaluate the integral:

$\int\frac{1}{cosx(sinx+2cosx)}dx$

Solution:

Let $I=\int\frac{1}{cosx(sinx+2cosx)}dx\\ =\int\frac{1}{sinxcosx+2cos^2x}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{sec^2x}{tanx + 2}dx$

Now, let us assume 2 + tanx = t

So, sec²x dx = dt

I = ∫dt/t

= log|t| + c

I = log|2 + tanx| + c

Question 10. Evaluate the integral:

$\int\frac{1}{sin^2x+sin2x}dx$

Solution:

Let $I=\int\frac{1}{sin^2x+sin2x}dx\\ =\int\frac{1}{sin^2x+2sinxcosx}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{sec^2x}{tan^2x+2tanx}dx$

Now, let us assume tanx = t

So, sec²x dx = dt

$=\int\frac{dt}{t^2+2t+(1)^2-(1)^2}\\ =\int\frac{dt}{(t+1)^2-(1)^2}\\ =\frac{1}{2}\log\left|\frac{t+1-1}{t+1+1}\right|+c\\ =\frac{1}{2}\log\left|\frac{t}{t+2}\right|+c\\ I=\frac{1}{2}\log\left|\frac{tanx}{tanx+2}\right|+c$

Question 11. Evaluate the integral:

$\int\frac{1}{cos2x+3sin^2x}dx$

Solution:

Let $I=\int\frac{1}{cos2x+3sin^2x}dx\\ =\int\frac{1}{2cos^2x-1+3sin^2x}dx$

On dividing numerator and denominator by cos²x, we get

$I=\int\frac{sec^x}{2-sec^2x+3tan^2x}dx\\ =\int\frac{sec^2x}{2-(1+tan^2x)^2+3tan^2x}dx\\ =\int\frac{sec^2x}{2-1-tan^2x+3tan^2x}dx\\ =\int\frac{dt}{1+2tan^2x}$

Now, let us assume √2tanx = t

So, √2sec²dx = dt

I = 1/√2 ∫1/(1 + t²)

= 1/√2 tan^-1t + c

Hence, I = 1/√2 tan^-1(√2tanx) + c

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

RD Sharma Class 12 Ex 19.22 Solutions Chapter 19 Indefinite Integrals

Question 1. Evaluate the integral:

Question 2. Evaluate the integral:

Question 3. Evaluate the integral:

Question 4. Evaluate the integral:

Question 5. Evaluate the integral:

Question 6. Evaluate the integral:

Question 7. Evaluate the integral:

Question 8. Evaluate the integral:

Question 9. Evaluate the integral:

Question 10. Evaluate the integral:

Question 11. Evaluate the integral:

Related Posts

Leave a Comment Cancel Reply