Integrate the function: cos ^{3} x e^{log sin | vedclass

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

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

Vedclass · Answer

(A) Given integral is $I = \int \cos ^{3} x e^{\log \sin x} dx$.Using the property $e^{\log f(x)} = f(x)$,we have $e^{\log \sin x} = \sin x$.So,the integral becomes $I = \int \cos ^{3} x \sin x dx$.Let $\cos x = t$. Then,differentiating both sides with respect to $x$,we get $-\sin x dx = dt$,or $\sin x dx = -dt$.Substituting these into the integral,we get $I = \int t^{3} (-dt) = -\int t^{3} dt$.Integrating $t^{3}$ with respect to $t$,we get $I = -\frac{t^{4}}{4} + C$.Substituting $t = \cos x$ back,we get $I = -\frac{\cos ^{4} x}{4} + C$.

Integrate the function: $\cos ^{3} x e^{\log \sin x}$

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