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(C) Let $I = \int \sin(101x) \sin^{99}x \, dx$. Using the identity $\sin(101x) = \sin(100x + x) = \sin(100x)\cos x + \cos(100x)\sin x$,we get: $I = \i

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(C) Let $I = \int \sin(101x) \sin^{99}x \, dx$. Using the identity $\sin(101x) = \sin(100x + x) = \sin(100x)\cos x + \cos(100x)\sin x$,we get: $I = \int (\sin(100x)\cos x + \cos(100x)\sin x) \sin^{99}x \, dx$ $I = \int \sin(100x) \cos x \sin^{99}x \, dx + \int \cos(100x) \sin^{100}x \, dx$. Applying integration by parts to the first integral,let $u = \sin(100x)$ and $dv = \cos x \sin^{99}x \, dx$. Then $du = 100 \cos(100x) \, dx$ and $v = \frac{\sin^{100}x}{100}$. $I = \sin(100x) \cdot \frac{\sin^{100}x}{100} - \int 100 \cos(100x) \cdot \frac{\sin^{100}x}{100} \, dx + \int \cos(100x) \sin^{100}x \, dx$. $I = \frac{\sin(100x) \sin^{100}x}{100} - \int \cos(100x) \sin^{100}x \, dx + \int \cos(100x) \sin^{100}x \, dx$. $I = \frac{\sin(100x) \sin^{100}x}{100} + C$. Comparing this with the given form $\frac{\sin(100x) \sin^{\lambda}x}{\mu} + C$,we have $\lambda = 100$ and $\mu = 100$. Therefore,$\frac{\lambda}{\mu} = \frac{100}{100} = 1$.

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

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