$\int \sin 101x{\sin }^{99}xdx=\frac{\sin \left(100x\right)(\sin x{)}^2}{\mu }\text{ then }\frac{\lambda +\mu }{100}\text{ equal to }$

$\begin{array}{l}\int \sin (100+1)x-{\sin }^{99}x=\int (\sin 100x\cos x+\cos 100x\sin x){\sin }^{99}x\\ =\sin 100x\frac{{\sin }^{100}x}{100}-\int \frac{1}{100}{\sin }^{100}x\cos 100x\left(100\right)+\int \cos 100x{\sin }^{100}xdx\\ =\frac{\sin 100x\cdot {\sin }^{100}x}{100}\lambda =100,\mu =100\end{array}$