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AP EAPCET 20-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 78 of 160

Marks: +1, -0

If

\int\limits_{0}^{\pi} \log(\sin x) \, dx = 8k

\int\limits_{0}^{\pi/4} \log(1+\tan x) \, dx =

_______

$k$
$-k$
$\frac{k}{2}$
$4k$

Solution:

\int\limits_{0}^{\pi} \log(\sin x) \, dx = 8k

Let

I = \int\limits_{0}^{\pi/4} \log(1+\tan x) \, dx

= \int\limits_{0}^{\pi/4} \log(1+\tan\theta) \, d\theta

= \int\limits_{0}^{\pi/4} \log(1+\tan(\pi/4 - \theta)) \, d\theta

I = \int\limits_{0}^{\pi/4} \log\left(\frac{1+\tan\theta+1-\tan\theta}{1+\tan\theta}\right) d\theta

I = \int\limits_{0}^{\pi/4} \log 2 \, d\theta - \int\limits_{0}^{\pi/4} \log(1+\tan\theta) \, d\theta

I = \int\limits_{0}^{\pi/4} \log 2 \, d\theta - I

\Rightarrow 2I = (\log 2)\left(\frac{\pi}{4}\right)

I = \frac{(\log 2)\pi}{8} = \frac{-1}{4} \times \left(\frac{-\pi}{2} \log 2\right)

= -\frac{1}{8}(-\pi \log 2)

= \frac{-1}{8} \times 8k = -k

[ \because \int\limits_{0}^{\pi/2} \log \sin x \, dx = \frac{-\pi}{2} \log 2

\Rightarrow \int\limits_{0}^{\pi} \log \sin x = -\pi \log 2 ]

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