int_0^{frac{pi}{2}} frac{dx}{1+(cot x)^{101}} | vedclass

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(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot x)^{101}}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:$I = \int_0^{\frac{

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$\frac{\pi}{4}$

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(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot x)^{101}}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:$I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot(\frac{\pi}{2}-x))^{101}}$Since $\cot(\frac{\pi}{2}-x) = 	an x$,we have:$I = \int_0^{\frac{\pi}{2}} \frac{dx}{1+(	an x)^{101}} = \int_0^{\frac{\pi}{2}} \frac{dx}{1+\frac{1}{(\cot x)^{101}}} = \int_0^{\frac{\pi}{2}} \frac{(\cot x)^{101}}{1+(\cot x)^{101}} dx$.Adding the two expressions for $I$:$2I = \int_0^{\frac{\pi}{2}} \frac{1+(\cot x)^{101}}{1+(\cot x)^{101}} dx = \int_0^{\frac{\pi}{2}} 1 dx = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$.Therefore,$I = \frac{\pi}{4}$.

$\int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot x)^{101}} = $

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