Question

Evaluate: ${\int }_0^{\pi }x,\sin x,{\cos }^{4}xdx$.

Answer 1

${\int }_0^{\pi }xdx=[x^2/2{]}_0^{\pi }=\frac{{\pi }^2}{2}$

${\int }_0^{\pi }\sin xdx=[-\cos x{]}_0^{\pi }=0$

${\int }_0^{\pi }{\cos }^{4}xdx={\int }_0^{\pi }[2+\cos 4x+2\cos 2x]dx=[2x+(\sin 4x)/4+\sin 2x{]}_0^{\pi }=2\pi$