Question

${\int }_0^{\pi /4}co{s}^{2}xdx$

Answer 1

${\int }_0^{\pi }{\cos }^{3}xdx$

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

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

Let $t=\sin x\Rightarrow dt=\cos xdx$

$={\int }_0^{\pi }(1-t^2)dt$

$={[t-\frac{t^3}{3}]}_0^{\pi }$

$={[\sin x-\frac{{\sin }^{3}x}{3}]}_0^{\pi }$

$=[0-0]=0$