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AIEEE 2011 Solved Paper

Section: Mathematics

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Question : 77 of 91

Marks: +1, -0

The value of

\int\limits_{0}^{1} \; \frac{8 \log(1+x)}{1+x^2} dx

$\frac{\pi}{8} \log 2$
$\frac{\pi}{2} \log 2$
$\log 2$
$\pi \log 2$

Solution:

I = \int\limits_{0}^{1} \; \frac{8 \log(1+x)}{1+x^2} dx

put

x = \tan \theta

\therefore \; \frac{dx}{d\theta} = \sec^2 \theta \Rightarrow dx = \sec^2 \theta \, d\theta

\therefore I = 8 \int\limits_{0}^{\pi/4} \; \frac{\log(1+\tan \theta)}{1+\tan^2 \theta} \cdot \sec^2 \theta \, d\theta

I = 8 \int\limits_{0}^{\pi/4} \log(1+\tan \theta) \, d\theta \ldots (i)

= 8 \int\limits_{0}^{\pi/4} \log \left[ 1 + \tan\left( \frac{\pi}{4} - \theta \right) \right] d\theta

= 8 \int\limits_{0}^{\pi/4} \log \left[ 1 + \frac{1 - \tan \theta}{1 + \tan \theta} \right] d\theta

= 8 \int\limits_{0}^{\pi/4} \log \left[ \frac{2}{1 + \tan \theta} \right] d\theta

= 8 \int\limits_{0}^{\pi/4} \left[ \log 2 - \log(1 + \tan \theta) \right] d\theta

I = 8 \cdot (\log 2) [x]_{0}^{\pi/4} - 8

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

I = 8 \cdot \frac{\pi}{4} \cdot \log 2 - I[

From equation

(i)]

\Rightarrow 2I = 2 \pi \log 2

,

\therefore I = \pi \log 2

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