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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 57 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 57 Maths Textbook Solution.

Answers (1)

Answer:

I=\frac{1}{4} \tan ^{-1} \frac{(2 \tan x+1)}{2}+c

Given:

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

Hint:

To solve this statement we have to divide numerator and denominator by cos²θ.

Solution: 

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

Dividing numerator and denominator by cos²x.

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

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

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

\left[\because 2 \tan x+1=t, d t=2 \sec ^{2} x d x, \sec ^{2} x d x=\frac{d t}{2}\right.

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

I=\frac{1}{2} \times \frac{1}{2} \tan ^{-1} \frac{t}{2}+c

I=\frac{1}{4} \tan ^{-1} \frac{(2 \tan x+1)}{2}+c

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