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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Evaluate :
$\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$
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Answer

$< I =\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$
$ =\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$
Adding eqns $(1)$ and $(2)$
$2 I =\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\cos x \sin x}{\sin ^4 x+\cos ^4 x} d x$
$\Rightarrow I =\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\cos x \sin x}{\sin ^4 x+\cos ^4 x} d x$
$ =\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x \sec ^2 x}{1+\tan ^4 x} d x$
$\text { Putting } \tan ^2 x=t \Rightarrow 2 \tan x \sec ^2 x d x=d t$
$\qquad \therefore I =\frac{\pi}{8} \int_0^{\infty} \frac{1}{1+t^2} d t$
$ \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)$
$=\frac{\pi^2}{16}$

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