Question

Evaluate: ${\int }_{\pi /4}^{\pi /2}\cos 2x\log \sin xdx$

Answer 1

${\int }_{\pi /4}^{\pi /2}\cos 2xln\sin xdx={\left[ln\sin x\int \cos 2xdx\right]}_{\pi /4}^{\pi /2}-{\int }_{\pi /4}^{\pi /2}\frac{\cos x}{\sin x}\left(\int {\cos }^{2}xdx\right)dx$

$={\left[ln\sin x\frac{\sin 2x}{2}\right]}_{\pi /4}^{\pi /2}-{\int }_{\pi /4}^{\pi /2}\frac{\cos x}{\sin x}\frac{\sin 2x}{2}dx$

$=-ln1/\sqrt{2}\frac{1}{2}-{\int }_{\pi /4}^{\pi /2}\frac{\cos x}{\sin x}\frac{2\sin x}{2}\cos xdx$

$\Rightarrow \frac{-ln\sqrt{2}}{2}-{\int }_{\pi /4}^{\pi /2}{\cos }^{2}xdx$

$\frac{ln\sqrt{2}}{2}-{\int }_{\pi /4}^{\pi /2}\frac{\cos 2x+1}{2}$

$\Rightarrow \frac{ln\sqrt{2}}{2}-\frac{1}{2}{[\frac{\sin 2x}{2}+x]}_{\pi /4}^{\pi /2}$

$\Rightarrow \frac{ln\sqrt{2}}{2}-\frac{1}{2}(\frac{\pi }{2}-\frac{1}{2}-\frac{\pi }{4})$.