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UPSC NDA II 2014 Mathematics Paper

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For the next three (03) items that follow:

Consider

I=\int\limits_{0}^{\pi}\frac{xdx}{1+\sin x}

[NDA II 2014]

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Question : 76 of 120

Marks: +1, -0

What is I equal to ?

$-\pi$
0
π
2π

Solution:

I=\int\limits_{0}^{\pi}\frac{x\,dx}{1+\sin x}\Rightarrow I=\int\limits_{0}^{\pi}\frac{(\pi-x)\,dx}{1+\sin x}

\Rightarrow I+I=\int\limits_{0}^{\pi}\frac{\pi\,dx}{1+\sin x}

\Rightarrow I=\frac{\pi}{2}\int\limits_{0}^{\pi}\frac{1-\sin x}{1-\sin^2 x}

\Rightarrow I=\frac{\pi}{2}\int\limits_{0}^{\pi}(\sec^2 x-\sec x\tan x)\,dx

\Rightarrow I=\frac{\pi}{2}[\tan x-\sec x]^{\pi}_0=\frac{\pi}{2}[0+1-0+1]=\pi

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