int cos^{-3/7} x sin^{-11/7} x , dx = | vedclass

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(C) Let $I = \int \cos^{-3/7} x \sin^{-11/7} x \, dx$. Here,$m = -3/7$ and $n = -11/7$. $m + n = -3/7 - 11/7 = -14/7 = -2$,which is a negative even in

Vedclass · Accepted Answer

$\frac{-7}{4} 	an^{-4/7} x + c$

Vedclass · Answer

(C) Let $I = \int \cos^{-3/7} x \sin^{-11/7} x \, dx$. Here,$m = -3/7$ and $n = -11/7$. $m + n = -3/7 - 11/7 = -14/7 = -2$,which is a negative even integer. We can rewrite the integral as: $I = \int \cos^{-3/7} x \sin^{-11/7} x \, dx = \int \left( \frac{\cos x}{\sin x} ight)^{-3/7} \sin^{-3/7 - 11/7} x \, dx$ $I = \int \cot^{-3/7} x \sin^{-2} x \, dx = \int \cot^{-3/7} x \csc^2 x \, dx$. Let $\cot x = t$,then $-\csc^2 x \, dx = dt$,or $\csc^2 x \, dx = -dt$. Substituting these into the integral: $I = \int t^{-3/7} (-dt) = -\int t^{-3/7} \, dt$. $I = -\frac{t^{-3/7 + 1}}{-3/7 + 1} + c = -\frac{t^{4/7}}{4/7} + c = -\frac{7}{4} t^{4/7} + c$. Since $t = \cot x$,we have $I = -\frac{7}{4} \cot^{4/7} x + c$. Using $\cot x = \frac{1}{	an x}$,we get $I = -\frac{7}{4} (	an x)^{-4/7} + c = -\frac{7}{4} 	an^{-4/7} x + c$. Thus,option $(c)$ is correct.

$\int \cos^{-3/7} x \sin^{-11/7} x \, dx = $

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