If int sin (101 x)(sin x)^{99} d x=frac{sin ( | vedclass

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(A) We are given the integral $I = \int \sin (101 x)(\sin x)^{99} d x$. Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we have $\sin(

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

If $\int \sin (101 x)(\sin x)^{99} d x=\frac{\sin (100 x)(\sin x)^\lambda}{\mu}+c$ then $\frac{\lambda}{\mu}=$

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