The integral _0^ 1 2 , ( 1 + 2x ) 1 + 4 x^2 dx , equals

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MCQ

The integral $\int\limits_0^{\frac{1}{2}} {\frac{{\ln \,\left( {1 + 2x} \right)}}{{1 + 4{x^2}}}} dx$ , equals

A
$\frac{\pi }{4}\,\ln \,2$
B
$\frac{\pi }{8}\,\ln \,2$
✓
$\frac{\pi }{16}\,\ln \,2$
D
$\frac{\pi }{32}\,\ln \,2$

✓

Answer

Correct option: C.

$\frac{\pi }{16}\,\ln \,2$

c
Let $I = \int\limits_0^{1/2} {\frac{{\ln (1 + 2x)}}{{\ln (1 + 2x)}}} dx$

or $\int\limits_0^{1/2} {\frac{{\ln (1 + 2x)}}{{1 + {{(2x)}^2}}}} dx$

Put $2 x=\tan \theta$

$\therefore \frac{2 d x}{d \theta}=\sec ^{2} \theta$ or $d x=\frac{\sec ^{2} \theta d \theta}{2}$

also when $x=0 \Rightarrow \theta=0$

and when $x=\frac{1}{2} \Rightarrow \theta=45^{\circ} \mathrm{or} \frac{\pi}{4}$

$\therefore I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\ln (1 + \tan \theta )}}{{1 + {{\tan }^2}\theta }}} \times \frac{{{{\sec }^2}\theta d\theta }}{2}$

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\frac{{\ln (1 + \tan \theta )}}{{1 + {{\tan }^2}\theta }}} \times {\sec ^2}\theta d\theta $

$\left(\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right)$

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } (1 + \tan \theta )d\theta $

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } \left[ {1 + \frac{{\tan \frac{\pi }{4} - \tan \theta }}{{1 + \tan \frac{\pi }{4} \times \tan \theta }}} \right]d\theta $

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } \left[ {1 + \frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right]d\theta $

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } \left[ {\frac{{1 + \tan \theta + 1 - \tan \theta }}{{1 + \tan \theta }}} \right]d\theta $

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } \left[ {\frac{2}{{1 + \tan \theta }}} \right]d\theta $

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } 2 - In\left[ {1 + \tan \theta } \right]d\theta $

$I = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } 2 \cdot d\theta - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\ln } (1 + \tan \theta )d\theta $

${\rm{I}} = \left. {\frac{1}{2}\ln 2\theta } \right|_0^{\pi /4} - I$

(from eq $(1))$

$\mathrm{I}+\mathrm{I}=\frac{1}{2} \ln 2\left(\frac{\pi}{4}-0\right)$

$2 \mathrm{I}=\frac{1}{2} \times \frac{\pi}{4} \times \ln 2$

$21=\frac{\pi}{8} \ln 2$ or $I=\frac{\pi}{16} \ln 2$