_0^ / 2 x x x ^4 x+ ^4 x d x= - Vidyadip

MCQ

$\int_0^{\pi / 2} \frac{x \sin x \cos x}{\cos ^4 x+\sin ^4 x} d x=$

A
$0$
B
$\frac{\pi}{8}$
C
$\frac{\pi^2}{8}$
✓
$\frac{\pi^2}{16}$

✓

Answer

Correct option: D.

$\frac{\pi^2}{16}$

(D)
Let $I =\int_0^{\pi / 2} \frac{x \sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$ ...(i)
$\therefore \quad I=\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\sin ^4 x+\cos ^4 x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\frac{\pi}{2} \int_0^{\pi / 2} \frac{\cos x \sin x}{\cos ^4 x+\sin ^4 x} d x$
$\Rightarrow I =\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan x \sec ^2 x}{1+\tan ^4 x} d x$
Put $\tan ^2 x= t \Rightarrow \tan x \sec ^2 x d x=\frac{ dt }{2}$
$\therefore \quad I=\frac{\pi}{8} \int_0^{\infty} \frac{ dt }{1+ t ^2}=\frac{\pi}{8}\left[\tan ^{-1} t \right]_0^{\infty}=\frac{\pi^2}{16}$

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