Question

Evaluate the integral ${\int }_0^{\frac{\pi }{4}}\log \left(\frac{\sin x+\cos x}{\cos x}\right)dx$

• A. $\frac{\pi }{4}\log 2$
• B. $\frac{\pi }{2}\log 2$
• C. $\frac{\pi }{8}\log 2$
• D. $\log 2$

Answer 1

Answer: (C)

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

Let $I={\int }_0^{\frac{\pi }{4}}\log \left(\frac{\sin x+\cos x}{\cos x}\right)dx={\int }_0^{\frac{\pi }{4}}\ln (1+\tan x)dx$........(1)

Now using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^{\frac{\pi }{4}}\ln (1+\tan (\frac{\pi }{4}-x\left)\right)dx={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1-\tan x}{1+\tan x})dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{2}{1+\tan x}\right)dx={\int }_0^{\frac{\pi }{4}}\ln 2dx-{\int }_0^{\frac{\pi }{4}}\ln (1+x)dx$.........(2)

Now add (1) and (2)

$\Rightarrow 2I=\ln 2{\int }_0^{\frac{\pi }{4}}dx=\left(\ln 2\right)[x{]}_0^{\frac{\pi }{4}}=\frac{\pi }{8}\ln 2$

$\therefore I=\frac{\pi }{8}\ln 2$