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NDA Math Definite Integrals

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Directions Next 2 Questions : Read the following information and answer the two items that follow:

Consider the integrals

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

and

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

[NDA II 2019]

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Question : 15 of 62

Marks: +1, -0

What is the value of

I_1

0
$\pi/2$
$\pi$
$2\pi$

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}} \left( \sec^{2}x - \tan x \sec x \right) dx

\Rightarrow 2I_{1} = 2\pi \left[ \tan x - \sec x \right]_{0}^{\pi/2}

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

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

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

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