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Question:

$\int_{0}^{\frac{x}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$

Solution:

Let $I=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \frac{\frac{(\sin x \cos x)}{\cos ^{4} x}}{\frac{\left(\cos ^{4} x+\sin ^{4} x\right)}{\cos ^{4} x}} d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{1+\tan ^{4} x} d x$

Let $\tan ^{2} x=t \Rightarrow 2 \tan x \sec ^{2} x d x=d t$

When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=1$

$\begin{aligned} \therefore I &=\frac{1}{2} \int_{0}^{t} \frac{d t}{1+t^{2}} \\ &=\frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1} \\ &=\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\ &=\frac{1}{2}\left[\frac{\pi}{4}\right] \\ &=\frac{\pi}{8} \end{aligned}$

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