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

Question

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

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Answer

$ \text { Let } I =\int_0^{\frac{\pi}{2}} \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x \quad \ldots \ldots . . \text { (i) }$
$\left.=\int_0^{\frac{\pi}{2}} \frac{\sin ^4\left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} \ldots \ldots . . \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore I =\int_0^{\frac{\pi}{2}} \frac{\cos ^4 x}{\cos ^4 x+\sin ^4 x} d x \ldots \ldots . . \text { (ii) } $
Adding (i) and (ii), we get
$ 2 I =\int_0^{\frac{\pi}{2}} \frac{\sin ^4 x}{\sin ^4 x+\cos ^4 x} d x+\int_0^{\frac{\pi}{2}} \frac{\cos ^4 x}{\cos ^4 x+\sin ^4 x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\sin ^4 x+\cos ^4 x}{\sin ^4 x+\cos ^4 x} d x$
$\therefore 2 I =\int_0^{\frac{\pi}{2}} 1 \cdot d x$
$\therefore I =\frac{1}{2}[x]_0^{\frac{\pi}{2}}$
$=\frac{1}{2}\left(\frac{\pi}{2}-0\right)$
$\therefore I =\frac{\pi}{4} $

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