If int sin ^5 x , dx = frac{-cos ^5 x}{5} + a | vedclass

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(A) We have the integral $I = \int \sin^5 x \, dx = \int \sin^4 x \cdot \sin x \, dx$. Using the identity $\sin^2 x = 1 - \cos^2 x$,we get $\sin^4 x =

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$-\frac{1}{3}$

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(A) We have the integral $I = \int \sin^5 x \, dx = \int \sin^4 x \cdot \sin x \, dx$. Using the identity $\sin^2 x = 1 - \cos^2 x$,we get $\sin^4 x = (1 - \cos^2 x)^2 = 1 - 2\cos^2 x + \cos^4 x$. Substituting this into the integral: $I = \int (1 - 2\cos^2 x + \cos^4 x) \sin x \, dx$. Let $u = \cos x$,then $du = -\sin x \, dx$,so $\sin x \, dx = -du$. $I = -\int (1 - 2u^2 + u^4) \, du = -[u - \frac{2u^3}{3} + \frac{u^5}{5}] + C = -\frac{u^5}{5} + \frac{2u^3}{3} - u + C$. Substituting $u = \cos x$ back: $I = -\frac{\cos^5 x}{5} + \frac{2}{3} \cos^3 x - \cos x + C$. Comparing this with the given expression $\frac{-\cos^5 x}{5} + a \cos^3 x + b \cos x + c$,we find $a = \frac{2}{3}$ and $b = -1$. Therefore,$a + b = \frac{2}{3} - 1 = -\frac{1}{3}$.

If $\int \sin ^5 x \, dx = \frac{-\cos ^5 x}{5} + a \cos ^3 x + b \cos x + c$,then $a + b =$

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