Question

Evaluate:${\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx$

Answer 1

${\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}2{\sin }^{2}xdx$

We know that $\cos 2x=1-2{\sin }^{2}x\Rightarrow 2{\sin }^{2}x=1-\cos 2x$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}(1-\cos 2x)dx$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}dx-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\cos 2xdx$

$=\frac{1}{2}{\left[x\right]}_0^{\frac{\pi }{2}}-\frac{1}{2}{\left[\frac{-\sin 2x}{2}\right]}_0^{\frac{\pi }{2}}$

$=\frac{1}{2}{\left[x\right]}_0^{\frac{\pi }{2}}-\frac{1}{4}{[-\sin 2x]}_0^{\frac{\pi }{2}}$

$=\frac{1}{2}[\frac{\pi }{2}-0]+\frac{1}{4}[\sin \pi -\sin 0]$

$=\frac{\pi }{4}+0$

$=\frac{\pi }{4}$