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

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Question

(A) Given the integral $I = \int \cos ^3 x \cdot e^{\log (\sin x)} d x$. Since $e^{\log (f(x))} = f(x)$,the integral simplifies to: $I = \int \cos ^3

Vedclass · Accepted Answer

$\frac{-\cos ^4 x}{4}+c$

Vedclass · Answer

(A) Given the integral $I = \int \cos ^3 x \cdot e^{\log (\sin x)} d x$. Since $e^{\log (f(x))} = f(x)$,the integral simplifies to: $I = \int \cos ^3 x \cdot \sin x d x$. Let $u = \cos x$. Then $du = -\sin x d x$,which implies $\sin x d x = -du$. Substituting these into the integral: $I = \int u^3 (-du) = -\int u^3 du$. Integrating $u^3$ gives $\frac{u^4}{4}$. Thus,$I = -\frac{u^4}{4} + c$. Substituting back $u = \cos x$,we get: $I = -\frac{\cos ^4 x}{4} + c$.

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

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