$\int \sin (101 x) \sin ^{99} x d x$ equals

$\frac{1}{100} \sin (100 x)(\sin x)^{100}+C$

Let $I=\int \sin (101 x) \sin ^{99} x d x$. Then,

$I =\int \sin (100 x+x) \sin ^{99} x d x$

$\Rightarrow I =\int \sin (100 x) \cos x \sin ^{99} x d x + \int \cos (100 x) \sin ^{100} x d x$

$\Rightarrow I=\sin (100 x) \frac{(\sin x)^{100}}{100}-\int \cos (100 x)(\sin x)^{100} x d x + \int \cos (100 x)(\sin x)^{100} d x+C$

$\Rightarrow I=\frac{1}{100} \sin (100 x)(\sin x)^{100}+C$