Question

The value of ${\int }_0^{2\pi }{\sin }^{100}x{\cos }^{99}x\mathrm{dx}$is equal to

Answer 1

We have,
$\mathrm{Let}f\left(x\right)={\sin }^{100}x{\cos }^{99}xf\left(2\pi -x\right)={\sin }^{100}\left(2\pi -x\right){\cos }^{99}\left(2\pi -x\right)\Rightarrow f(2\pi -x)={\sin }^{100}x{\cos }^{99}x=f\left(x\right)\therefore {\int }_0^{2\pi }{\sin }^{100}x{\cos }^{99}x\mathrm{dx}=2{\int }_0^{\pi }{\sin }^{100}x{\cos }^{99}x\mathrm{dx}\left(\mathrm{As}{\int }_0^{2a}f\left(x\right)\mathrm{dx}=2{\int }_0^{a}f\left(x\right)\mathrm{dx},\mathrm{if}f(2a-x)=f\left(x\right)\right]\mathrm{Let}I=2{\int }_0^{\pi }{\sin }^{100}x{\cos }^{99}x\mathrm{dx}=2{\int }_0^{\pi }{\sin }^{100}\left(\pi -x\right){\cos }^{99}\left(\pi -x\right)\mathrm{dx}=-2{\int }_0^{\pi }{\sin }^{100}x{\cos }^{99}x\mathrm{dx}=-I.I=-I\Rightarrow 2I=0\Rightarrow I=0$