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(C) We need to evaluate the integral $I = \int \sin^3 x \, dx$. Using the identity $\sin^2 x = 1 - \cos^2 x$,we can rewrite the integral as: $I = \int

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$\frac{\cos^3 x}{3} - \cos x + C$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int \sin^3 x \, dx$. Using the identity $\sin^2 x = 1 - \cos^2 x$,we can rewrite the integral as: $I = \int \sin^2 x \cdot \sin x \, dx = \int (1 - \cos^2 x) \sin x \, dx$. Let $u = \cos x$,then $du = -\sin x \, dx$,which implies $\sin x \, dx = -du$. Substituting these into the integral: $I = \int (1 - u^2) (-du) = \int (u^2 - 1) \, du$. Integrating with respect to $u$: $I = \frac{u^3}{3} - u + C$. Substituting $u = \cos x$ back into the expression: $I = \frac{\cos^3 x}{3} - \cos x + C$.

$\int \sin^3 x \, dx$ is equal to

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