Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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