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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 41 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 41 Maths Textbook Solution.

Answers (1)

Answer:

\tan ^{-1}\left(\tan ^{2} x\right)+c

Given:

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

Hint:

In this statement we have to convert sin in term of tan.

Solution: 

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

I=\frac{2 \sin x \cos x}{\cos ^{4} x\left(\tan ^{4} x+1\right)} d x

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

I=2 \int \frac{\tan x \sec ^{2} x}{\tan ^{4} x+1} d x                        \left[\because \operatorname{let} \tan x=t, \sec ^{2} x d x=d t\right]

I=2 \int \frac{t d t}{t^{4}+1}

I=\int \frac{d y}{y^{2}+1}\left[\because \mathrm{t}^{2}=\mathrm{y}, 2 \mathrm{tdt}=\mathrm{dy}\right]

    =\tan ^{-1} y+c

    =\tan ^{-1}\left(t^{2}\right)+c

     =\tan ^{-1}\left(\tan ^{2} x\right)+c

 

 

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