$\begin{aligned} & \int \sin (101 x) \cdot(\sin x)^{99} d x \\ & =\int \sin (100 x+x) \cdot(\sin x)^{99} d x \\ & =\int(\sin 100 x \cdot \cos x+\cos 100 x \cdot \sin x)(\sin x)^{99} d x \\ & =\iint_{\text {I }} \sin (100 x) \cdot \underbrace{\cos x \cdot(\sin x)^{99}}_{\text {II }} d x+\int \cos 100 x \cdot(\sin x)^{100} d x \\ & \because \int \cos x \cdot(\sin x)^{99} d x \\ & \quad \text { Let } \sin x=t \\ & \quad \cos x d x=d t \\ & \quad \int t^{99} d t=\frac{t^{100}}{100}=+\frac{(\sin x)^{100}}{100} \\ & =(\sin 100 x) \cdot \frac{(\sin x)^{100}}{100}+\int-100 \cos 100 x \cdot \frac{(\sin x)^{100}}{100} d x \\ & =(\sin 100 x) \cdot \frac{(\sin x)^{100}}{100}-\int(\cos 100 x) \cdot(\sin x)^{100} d x \\ & +\int(\cos 100 x) \cdot(\sin x)^{100} d x\end{aligned}$
$\begin{aligned} & =\frac{\sin (100 x) \cdot(\sin x)^{100}}{100}+C \\ & \therefore \quad \lambda=100, \mu=100 \\ & \therefore \quad \frac{\lambda}{\mu}=1 .\end{aligned}$