Question

Evaluate the integral

${\int }_0^{1}co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$

• A. $\frac{\pi }{2}-\log 2$
• B. $\frac{\pi }{2}+\log 2$
• C. $\frac{\pi }{4}$ - log 2
• D. $\frac{\pi }{4}$ - log 3

Answer 1

The correct option is A $\frac{\pi }{2}-\log 2$

${\int }_0^{1}co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$x=tan\theta$

$dx=se{c}^2\theta \cdot d\theta$

${\int }_0^{\frac{\pi }{4}}co{s}^{-1}\left(cos2\theta \right)co{s}^2\theta d\theta$

${\int }_0^{\frac{\pi }{4}}2\theta se{c}^2\theta d\theta =2{\int }_0^{\frac{\pi }{4}}\theta se{c}^2\theta d\theta$

$2[\theta tan\theta +log|cos\theta |{]}_0^{\frac{\pi }{4}}$

$2\left[\right(\frac{\pi }{4}-0)+log(\frac{1}{\sqrt{2}}\left)\right]$

$\frac{\pi }{2}-2log\left(\sqrt{2}\right)=\frac{\pi }{2}-\frac{2}{2}log\left(2\right)$

$=\frac{\pi }{2}-log2$