#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.22 Question 8 Maths Textbook Solution.

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

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

Hint: Divide Numerator and denominator by $\cos ^{4} x$

Solution:

$\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x$                        $(\sin 2 x=2 \sin x \cos x)$

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

On dividing Numerator and denominator by $\cos ^{4} x$

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

$Let$

\begin{aligned} &\tan ^{2} x=t \\ &2 \tan x \sec ^{2} x d x=d t \end{aligned}                            (Differentiate w.r.t x)

$Now,$$\int \frac{1}{t^{2}+1} d t$

$=\tan ^{-1} t+c$                                                              $\left[\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+c\right]$

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

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