$\int (\sin \left(101x\right).{\sin }^{99}x)dx$  equals
 

$$\begin{gathered} I=\int \sin (100x+x).{\left(\sin x\right)}^{99}dx \\ =\int (\sin \left(100x\right)\cos x+\cos 100x.\sin x){\left(\sin x\right)}^{99}dx \\ =\int \underset{I}{\underbrace{\sin \left(100x\right)}}\underset{II}{\underbrace{\cos x.{\left(\sin x\right)}^{99}}}dx +\int \cos \left(100x\right).{\left(\sin x\right)}^{100}dx \\ =\frac{\sin \left(100x\right){\left(\sin x\right)}^{100}}{100}-\frac{100}{100}\int \cos \left(100x\right).{\left(\sin x\right)}^{100}dx +\int \cos \left(100x\right).{\left(\sin x\right)}^{100}dx \\ =\frac{\sin \left(100x\right){\left(\sin x\right)}^{100}}{100}+C \end{gathered}$$