Evaluate : _0^ 2 x x x ^4 x+ ^4 x d x - Vidyadip

Question

Evaluate :
$
\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x
$

✓

Answer

<$
\begin{aligned}
I & =\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x \\
& {\left(\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right) } \\
& \int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x
\end{aligned}
$
Adding eqns (1) and (2)
$
\begin{aligned}
2 I & =\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\cos x \sin x}{\sin ^4 x+\cos ^4 x} d x \\
\Rightarrow \quad I & =\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\cos x \sin x}{\sin ^4 x+\cos ^4 x} d x \\
& =\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x \sec ^2 x}{1+\tan ^4 x} d x
\end{aligned}
$
$\begin{array}{l}\text { Putting } \tan ^2 x=t \Rightarrow 2 \tan x \sec ^2 x d x=d t \\ \qquad \begin{aligned} \therefore \quad I & =\frac{\pi}{8} \int_0^{\infty} \frac{1}{1+t^2} d t \\ & \quad\left(\because x=0 \text { then } t=0 ; x=\frac{\pi}{2} \text { then } t=\infty\right) \\ & =\frac{\pi}{8}\left(\tan ^{-1} t\right)_0^{\infty} \\ & =\frac{\pi}{8}\left(\tan ^{-1}(\infty)-\tan ^{-1}(0)\right) \\ & =\frac{\pi}{8}\left(\frac{\pi}{2}-0\right)\end{aligned}\end{array}$
$=\frac{\pi^2}{16}$

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