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TS EAMCET 2015 Solved Paper

Section: Mathematics

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Question : 5 of 160

Marks: +1, -0

\int\limits_{0}^{\frac{\pi}{2}} \frac{16x \sin x \cos x \, dx}{\sin^4 x + \cos^4 x} =

$\frac{\pi^2}{4}$
$\frac{\pi^2}{2}$
$\pi^2$
$2\pi^2$

Solution:

Let

I = \int\limits_{0}^{\frac{\pi}{2}} \; \frac{16x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx

= \int\limits_{0}^{\frac{\pi}{2}} \; \frac{16\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)} \, dx

= \int\limits_{0}^{\frac{\pi}{2}} \; \frac{(8\pi - 16x) \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx

On adding Eqs. (i) and (ii), we get

2I = 8\pi \int\limits_{0}^{\frac{\pi}{2}} \; \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx

\therefore

= 4\pi \int\limits_{0}^{\frac{\pi}{2}} \; \frac{2\sin x \cos x}{\sin^4 x + \cos^4 x} \, dx

I = 2\pi \int\limits_{0}^{\frac{\pi}{2}} \; \frac{2\tan x \sec^2 x}{\tan^4 x + 1} \, dx

= 2\pi \int\limits_{0}^{\frac{\pi}{2}} d\left\{ \tan^{-1}(\tan^2 x) \right\}

= 2\pi \left[ \tan^{-1}(\tan^2 x) \right]_{0}^{\frac{\pi}{2}} = 2\pi \left[ \frac{\pi}{2} - 0 \right]

\therefore \;\; I = \pi^2

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