If int cos ^{frac{3}{5}} x cdot sin ^3 x ,d x | vedclass

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(C) Let $I = \int \cos ^{\frac{3}{5}} x \sin ^3 x \,d x$. We can rewrite $\sin ^3 x$ as $\sin ^2 x \cdot \sin x = (1 - \cos ^2 x) \sin x$. So,$I = \in

Vedclass · Accepted Answer

$\left(\frac{8}{5}, \frac{18}{5}\right)$

Vedclass · Answer

(C) Let $I = \int \cos ^{\frac{3}{5}} x \sin ^3 x \,d x$. We can rewrite $\sin ^3 x$ as $\sin ^2 x \cdot \sin x = (1 - \cos ^2 x) \sin x$. So,$I = \int \cos ^{\frac{3}{5}} x (1 - \cos ^2 x) \sin x \,d x$. $I = \int (\cos ^{\frac{3}{5}} x - \cos ^{\frac{13}{5}} x) \sin x \,d x$. Let $\cos x = t$,then $-\sin x \,d x = dt$,or $\sin x \,d x = -dt$. Substituting these into the integral: $I = -\int (t^{\frac{3}{5}} - t^{\frac{13}{5}}) dt = -\int t^{\frac{3}{5}} dt + \int t^{\frac{13}{5}} dt$. Integrating with respect to $t$: $I = -\frac{t^{\frac{3}{5} + 1}}{\frac{3}{5} + 1} + \frac{t^{\frac{13}{5} + 1}}{\frac{13}{5} + 1} + c$. $I = -\frac{t^{\frac{8}{5}}}{\frac{8}{5}} + \frac{t^{\frac{18}{5}}}{\frac{18}{5}} + c$. Substituting $t = \cos x$ back: $I = -\frac{1}{\frac{8}{5}} \cos ^{\frac{8}{5}} x + \frac{1}{\frac{18}{5}} \cos ^{\frac{18}{5}} x + c$. Comparing this with the given expression $\frac{-1}{m} \cos ^{m} x + \frac{1}{n} \cos ^{n} x + c$,we find $m = \frac{8}{5}$ and $n = \frac{18}{5}$. Thus,$(m, n) = \left(\frac{8}{5}, \frac{18}{5}\right)$.

If $\int \cos ^{\frac{3}{5}} x \cdot \sin ^3 x \,d x = \frac{-1}{m} \cos ^{m} x + \frac{1}{n} \cos ^{n} x + c$,(where $c$ is the constant of integration),then $(m, n) = $

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