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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.

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


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


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


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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