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UPSC NDA 2 2019 Mathematics Solved Paper

Directions (Q. Nos. 94 and 95): Read the following information and answer the two items that follow:

Consider the integrals

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

and

I_{2}=\int\limits_{0}^{\pi}\frac{(\pi-x)\,dx}{1-\sin(\pi+x)}

[NDA II 2019]

Question : 94 of 120

Marks: +1, -0

What is the value of

I_1

Solution: 👈: Video Solution

I_1=\int\limits_{0}^{\pi}\;\;\frac{xdx}{1+\sin x}

....(i)

\Rightarrow I_1=\int\limits_{0}^{\pi}\frac{\pi-x}{1+\sin(\pi-x)}dx\left[\because\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx\right]

\Rightarrow I_1=\int\limits_{0}^{\pi}\;\;\frac{\pi-x}{1+\sin x}dx

....(ii)

Adding (i) and (ii), we get

\Rightarrow 2I_1=\int\limits_{0}^{\pi}\frac{x+\pi-x}{1+\sin x}dx

\Rightarrow 2I_1=\pi\int\limits_{0}^{\pi}\frac{1}{1+\sin x}dx

\left[\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}\left\{f(x)+f(2a-x)\right\}dx\right]

\Rightarrow 2I_1=\pi\int\limits_{0}^{\frac{\pi}{2}}\left[\frac{1}{1+\sin x}+\frac{1}{1+\sin(\pi-x)}\right]dx

\Rightarrow 2I_1=\pi\int\limits_{0}^{\frac{\pi}{2}}\left[\frac{1}{1+\sin x}+\frac{1}{1+\sin x}\right]dx

\Rightarrow 2I_1=2\pi\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin x}dx

\Rightarrow 2I_1=2\pi\int\limits_{0}^{\frac{\pi}{2}}\frac{1-\sin x}{1-\sin^{2}x}dx

\Rightarrow 2I_1=2\pi\int\limits_{0}^{\frac{\pi}{2}}(\sec^{2}x-\tan x\sec x)dx

\Rightarrow 2I_1=2\pi\left[\tan x-\sec x\right]_{0}^{\frac{\pi}{2}}

=2\pi\left[\frac{\sin x-1}{\cos x}\right]_{0}^{\frac{\pi}{2}}

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

2I_1=2\pi\left[\frac{-\cos^{2}x}{\cos x(1+\sin x)}\right]_{0}^{\frac{\pi}{2}}

=2\pi\left[\frac{-\cos x}{1+\sin x}\right]_{0}^{\frac{\pi}{2}}

=2\pi(0+1)=2\pi

2I_1=2\pi\therefore I_1=\pi\Rightarrow\int\limits_{0}^{\pi}\;\;\frac{xdx}{1+\sin x}=\pi

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