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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

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Question : 6 of 160

Marks: +1, -0

\int\limits_{0}^{\pi}\frac{xdx}{4\cos^{2}x+9\sin^{2}x}=

$\frac{\pi^{2}}{12}$
$\frac{\pi^{2}}{4}$
$\frac{\pi^{2}}{6}$
$\frac{\pi^{2}}{3}$

Solution:

Consider the integration,

I=\int\limits_{0}^{\pi}\;\;\frac{xdx}{4\cos^{2}x+9\sin^{2}x}.....(1)

The above integration can be simplified as,

I=\int\limits_{0}^{\pi}\;\;\frac{xdx}{4\cos^{2}x+9\sin^{2}x}

l=\int\limits_{0}^{\pi}\;\;\frac{(\pi-x)dx}{4\cos^{2}(\pi-x)+9\sin^{2}(\pi-x)}

=\int\limits_{0}^{\pi}\;\;\frac{(\pi-x)dx}{4\cos^{2}x+9\sin^{2}x}....(2)

Add equation (1) and (2).

2I=\int\limits_{0}^{\pi}\;\;\frac{\pi dx}{4\cos^{2}x+9\sin^{2}x}

=\int\limits_{0}^{\pi}\;\;\frac{\pi\sec^{2}xdx}{4+9\tan^{2}x}

=\int\limits_{0}^{\frac{\pi}{2}}\;\;\frac{2\pi\sec^{2}xdx}{4+9\tan^{2}x}

I=\;\;\frac{\pi}{9}\int\limits_{0}^{\frac{\pi}{2}}\;\;\frac{\sec^{2}xdx}{\frac{4}{9}+\tan^{2}x}

Put

\tan x=t

sec

{}^{2}x\,dx=dt

At

x=0, t=0

x=\frac{\pi}{2}, t=\infty

Then

I=\;\;\frac{\pi}{9}\int\limits_{0}^{\infty}\;\;\frac{dt}{\left(\frac{2}{3}\right)^{2}+t^{2}}

=\;\;\frac{\pi}{9}\times\;\;\frac{3}{2}\left[\tan^{-1}\frac{3t}{2}\right]_{0}^{\infty}

=\;\;\frac{\pi}{9}\times\;\;\frac{3}{2}\times\;\;\frac{\pi}{2}

=\;\;\frac{\pi^{2}}{12}

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