Question

Integrate:${\int }_0^{\pi /2}\sin x.\sin 2xdx$

Answer 1

${\int }_0^{\pi /2}sinxsin2xdx$

$=\frac{1}{2}{\int }_0^{\pi /2}2sinxsin2xdx$

$=\frac{1}{2}{\int }_0^{\pi /2}(cosx-cos3x)dx$

$=\frac{1}{2}({|}_0^{\pi /2}sinx-{\int }_0^{\pi /2})\frac{sin3x}{3}$

$=\frac{1}{2}(1-(\frac{-1}{3})$

$=\frac{2}{3}$