int sin^5 x , dx = | vedclass

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Question

(A) Let $I = \int \sin^5 x \, dx = \int \sin^4 x \cdot \sin x \, dx$. We can write $\sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2$. So,$I = \int (1 - \co

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$-\cos x + \frac{2}{3} \cos^3 x - \frac{\cos^5 x}{5} + c$,where $c$ is the constant of integration

Vedclass · Answer

(A) Let $I = \int \sin^5 x \, dx = \int \sin^4 x \cdot \sin x \, dx$. We can write $\sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2$. So,$I = \int (1 - \cos^2 x)^2 \sin x \, dx$. Let $u = \cos x$,then $du = -\sin x \, dx$,which implies $\sin x \, dx = -du$. Substituting these into the integral: $I = \int (1 - u^2)^2 (-du) = -\int (1 - 2u^2 + u^4) \, du$. Integrating term by term: $I = -(u - \frac{2u^3}{3} + \frac{u^5}{5}) + c = -u + \frac{2}{3} u^3 - \frac{1}{5} u^5 + c$. Substituting $u = \cos x$ back: $I = -\cos x + \frac{2}{3} \cos^3 x - \frac{1}{5} \cos^5 x + c$.

$\int \sin^5 x \, dx =$

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