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NDA Math Definite Integrals

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Consider the integral

I=\int\limits_{0}^{\pi}\ln(\sin x)\,dx

[NDA II 2014]

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Question : 57 of 62

Marks: +1, -0

What is

I=\int\limits_{0}^{\pi/2}\ln(\cos x)dx

I/2
I
2I
4I

Solution:

I=\int\limits_{0}^{\pi}\ln(\sin x)dx

\Rightarrow I=\int\limits_{0}^{\pi/2}\ln(\sin x)dx+\int\limits_{\pi/2}^{\pi}\ln(\sin x)dx

...(1)

Substitute

\pi - x = t

in the second integral

\Rightarrow dx = -dt

\Rightarrow \int\limits_{\pi/2}^{\pi}\ln(\sin x)dx = \int\limits_{\pi/2}^{0}\ln(\sin(\pi - t))(-dt)

= -\int\limits_{\pi/2}^{0}\ln(\sin t)dt = \int\limits_{0}^{\pi/2}\ln(\sin t)dt = \int\limits_{0}^{\pi/2}\ln(\sin x)dx

....(2)

From 1 and 2,

I=2\int\limits_{0}^{\pi/2}\ln(\sin x)dx

\Rightarrow \int\limits_{0}^{\pi/2}\ln(\sin x)dx = \frac{I}{2}

\Rightarrow \int\limits_{0}^{\pi/2}\ln(\sin(\pi/2 - x))dx = \frac{I}{2}

\Rightarrow \int\limits_{0}^{\pi/2}\ln(\cos x)dx = \frac{I}{2}

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