Question

Integrate : ${\int }_0^{\frac{\pi }{2}}\log \cos xdx$.

• A. $\frac{\pi }{2}\log 4$
• B. $-\frac{\pi }{2}\log 2$
• C. $\frac{\pi }{2}\log 2$
• D. None of these

Answer 1

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

Let $I={\int }_0^{\pi /2}log\left(cosx\right)dx$

By substituting $x$ for $\pi /2-x$ , we have

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

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

$={\int }_0^{\pi /2}log2dx+{\int }_0^{\pi /2}log\left(sin\frac{x}{2}\right)dx+{\int }_0^{\pi /2}log\left(cos\frac{x}{2}\right)dx$

using substitution $x=\frac{x}{2}$

$I=\frac{\pi }{2}log2+2{\int }_0^{\pi /2}log\left(sinx\right)dx+2{\int }_0^{\pi /2}log\left(cosx\right)dx$

$=\frac{\pi }{2}log2+I_1+I_2$

For $I_2$ , using substitution $x=\frac{\pi }{2}-x$

$I_2=2{\int }_{\pi /4}^{\pi /2}log\left(sinx\right)dx$

It gives that $I_1+I_2=2I$

$I=\frac{\pi }{2}log2+2I$

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

$\Rightarrow {\int }_0^{\pi /2}log\left(cosx\right)dx=\frac{-\pi }{2}log2$