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.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.22
CategoryRD Sharma Solutions

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/(t+ 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

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