Question

If $I_1=\underset{0}{\overset{2\pi }{\int }}\ln (1-{\tan }^2\frac{x}{8})dx,I_2=\underset{0}{\overset{2\pi }{\int }}\ln (1+{\tan }^2\frac{x}{8})dx$, then the value of $(I_1-I_2)$ is

• A. $-\pi \ln 2$
• B. $-\pi \ln 4$
• C. $-\frac{\pi }{2}\ln 8$
• D. $-\pi \ln 16$

Answer 1

The correct option is B $-\pi \ln 4$

$I_1-I_2=\underset{0}{\overset{2\pi }{\int }}\ln \left(\frac{1-{\tan }^2\frac{x}{8}}{1+{\tan }^2\frac{x}{8}}\right)dx\Rightarrow I_1-I_2=\underset{0}{\overset{2\pi }{\int }}\ln \left(\cos \frac{x}{4}\right)dx$
Putting $x=4y\Rightarrow dx=4dy$
$\Rightarrow I_1-I_2=4\underset{0}{\overset{\pi /2}{\int }}\ln \left(\cos y\right)dy$
We know that,
$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\ln \left(\cos x\right)dx=-\frac{\pi }{2}\ln 2\therefore I_1-I_2=-2\pi \ln 2=-\pi \ln 4$