Question
$\int {\left( {\sin \left( {101x} \right).{{\sin }^{99}}x} \right)} dx = \frac{{\sin \left( {100x} \right){{\left( {\sin x} \right)}^\lambda }}}{\mu } + C$ where $C$ is constant of integration then $\frac{\lambda }{\mu }$ is equal to

Answer

c
$\mathrm{I}=\int \sin (100 \mathrm{x}+\mathrm{x}) \cdot \sin ^{99} \mathrm{xd} \mathrm{x}=$

$\int(\sin 100 x \cos x+\cos 100 x \sin x) \cdot \sin ^{99} x d x$

$ = \int {\underbrace {\sin 100x}_I} \underbrace {\cos x{{\sin }^{99}}x}_{II}dx + \int {\cos } (100x){(\sin x)^{100}}dx$

$\Rightarrow I=\frac{\sin (100 x)(\sin x)^{100}}{100}-\frac{100}{100}$

$\int \cos (100 x)(\sin x)^{100} d x+\int \cos (100 x)(\sin x)^{100} d x$

$\Rightarrow \mathrm{I}=\frac{\sin (100 \mathrm{x})(\sin \mathrm{x})^{100}}{100}+\mathrm{c}$

$\lambda=100, \quad \mu=100 \Rightarrow \frac{\lambda}{\mu}=1$

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