Question

Let $u=\underset{0}{\overset{1}{\int }}\frac{\ln \left(x+1\right)}{x^2+1}dx$ and $v=\underset{0}{\overset{\pi /2}{\int }}\ln \left(\sin 2x\right)dx$ then -

• A. a) u = 4v
• B. b) 4u + v = 0
• C. c) u + 4v = 0
• D. d) 2u + v = 0

Answer 1

Answer: (B)

b) 4u + v = 0

Let $={\int }_0^1\frac{(x+1)}{x^2+1}dx$

Let $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$

$u={\int }_0^{\pi /4}\frac{\ln (1+\tan \theta )}{{\sec }^2\theta }{\sec }^2\theta d\theta$

$={\int }_0^{\pi /4}\ln (1+\tan \theta )d\theta$

$={\int }_0^{\pi /4}\ln (1+\tan \left(\frac{\pi }{4}\right))d\theta$

$={\int }_0^{\pi /4}\ln (1+\frac{1-\tan \theta }{1+\tan \theta })d\theta$

$={\int }_0^{\pi /4}\ln \left(\frac{2}{1+\tan \theta }\right)d\theta$

$={\int }_0^{\pi /4}\ln 2d\theta -={\int }_0^{\pi /4}\ln (1+\tan theta)d\theta$

$\Rightarrow 2u=\frac{\pi }{4}\ln 2$

$\Rightarrow u=\frac{\pi }{8}\ln 2$

Again

$v={\int }_0^{\pi /2}\ln \left(\sin 2x\right)dx$

$={\int }_0^{\pi /2}\left(2\sin x\cos x\right)dx$

$=\frac{\pi }{2}\ln 2+{\int }_0^{\pi /2}\ln \sin xdx+{\int }_0^{\pi /2}\ln \cos xdx$

Let $I={\int }_0^{\pi /2}\ln \sin xdx$

$I={\int }_0^{\pi /2}\log \left(2\sin \frac{\pi }{2}\cos \frac{\pi }{2}\right)dx$

$=\frac{\pi }{2}{\log }_2+{\int }_0^{\pi /2}\ln \sin \frac{x}{2}dx+{\int }_0^{\pi /2}\log \cos \frac{x}{2}dx$

$=\frac{\pi }{2}{\log }_2+I_1+I_2$

Put $\frac{x}{2}=t\Rightarrow dx=2dt$

$I=\frac{\pi }{2}{\log }_2+2{\int }_0^{\pi /4}\ln \sin tdt+2{\int }_0^{\pi /4}\ln \cos tdt$

Now put $t=\frac{\pi }{2}-t$ in ${\int }_0^{\pi /4}\ln \cos tdt$,

we get ${\int }_0^{\pi /4}\ln \cos tdt={\int }_{\pi /4}^{\pi /2}\ln \sin tdt$

$\therefore I=\ln 2+2{\int }_0^{\pi /4}\ln \sin tdt+2{\int }_{\pi /4}^{\pi /2}\ln \sin tdt$

$=\frac{\pi }{2}\ln 2+2{\int }_0^{\pi /2}\ln \sin tdt$

$\Rightarrow -I=\frac{\pi }{2}\ln 2\Rightarrow I=-\frac{\pi }{2}\ln 2$

$v=\frac{\pi }{2}\ln 2-\frac{\pi }{2}\ln 2-\frac{\pi }{2}\ln 2$

$v=-\frac{\pi }{2}\ln 2$

$\therefore u=\frac{\pi }{8}\ln 2,v=-\frac{\pi }{2}\ln 2$

$\therefore 4u+v=0$