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NDA Math Definite Integrals

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Consider

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

[NDA II 2014]

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Question : 55 of 62

Marks: +1, -0

What is

\int\limits_{0}^{\pi} \frac{dx}{1+\sin x}

1
2
4
$-2$

Solution:

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

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

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

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

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