int cos ^3 x e^{log (sin x)^2} d x= | vedclass

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Question

(B) Given integral is $I = \int \cos ^3 x e^{\log (\sin x)^2} dx$. Using the property $e^{\log f(x)} = f(x)$,we have $e^{\log (\sin x)^2} = (\sin x)^2

Vedclass · Accepted Answer

$\frac{\sin ^3 x}{3}-\frac{\sin ^5 x}{5}+c$

Vedclass · Answer

(B) Given integral is $I = \int \cos ^3 x e^{\log (\sin x)^2} dx$. Using the property $e^{\log f(x)} = f(x)$,we have $e^{\log (\sin x)^2} = (\sin x)^2$. So,$I = \int \cos ^3 x \sin ^2 x dx$. We can write this as $I = \int \cos ^2 x \sin ^2 x \cos x dx$. Since $\cos ^2 x = 1 - \sin ^2 x$,we get $I = \int (1 - \sin ^2 x) \sin ^2 x \cos x dx$. Let $\sin x = t$,then $\cos x dx = dt$. Substituting these into the integral,we get $I = \int (1 - t^2) t^2 dt = \int (t^2 - t^4) dt$. Integrating with respect to $t$,we get $I = \frac{t^3}{3} - \frac{t^5}{5} + c$. Substituting $t = \sin x$ back,we get $I = \frac{\sin ^3 x}{3} - \frac{\sin ^5 x}{5} + c$.

$\int \cos ^3 x e^{\log (\sin x)^2} d x=$

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