Question

Evaluate the following definite integral:

${\int }_{-\pi /4}^{\pi /4}\log (\cos x+\sin x)dx$

• A. $\pi$ log2
• B. $-\pi$ log2
• C. $-\frac{\pi }{4}\log 2$
• D. ${\pi }^2\log 2$

Answer 1

Answer: (C)

$-\frac{\pi }{4}\log 2$

Consider, $I={\int }_{-\pi /4}^{\pi /4}log(cosx+sinx)dx$

$I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}log\left[\sqrt{2}sin(\frac{\pi }{4}+x)\right]dx$

let, $\frac{\pi }{4}+x=t$ $\Rightarrow$ $dx=dt$

So, $x\to \frac{-\pi }{4}\Rightarrow t\to 0$ and $x\to \frac{\pi }{4}\Rightarrow t\to \frac{\pi }{2}$

$\Rightarrow$ $I={\int }_0^{\frac{\pi }{2}}\log \left(\sqrt{2}\sin t\right)dt$

$\Rightarrow$$I={\int }_0^{\frac{\pi }{2}}\log \left(\sqrt{2}\right)+{\int }_0^{\frac{\pi }{2}}\log \left(\sin t\right)dt$

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

$\Rightarrow$$I=\frac{\pi }{4}\log 2-\frac{\pi }{2}\log \left(2\right)$

$\Rightarrow$$I=-\frac{\pi }{4}\log 2$