Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.22 (original) (raw)

Last Updated : 3 Sep, 2024

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

Summary

Exercise 19.22 of RD Sharma's Class 12 Solutions focuses on advanced techniques for solving indefinite integrals. This exercise likely combines various methods such as integration by parts, trigonometric substitutions, partial fractions decomposition, and other specialized approaches for complex integrands. The problems are designed to challenge students' understanding of when and how to apply different integration techniques, often requiring a combination of methods to solve a single problem.