int sin^5 x cos^4 x , dx = | vedclass

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Question

(A) Let $I = \int \sin^5 x \cos^4 x \, dx$. Substitute $\cos x = t$,then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$. We can write $\sin^5 x$ as $(\s

Vedclass · Accepted Answer

$-\frac{1}{5}\cos^5 x + \frac{2}{7}\cos^7 x - \frac{1}{9}\cos^9 x + c$

Vedclass · Answer

(A) Let $I = \int \sin^5 x \cos^4 x \, dx$. Substitute $\cos x = t$,then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$. We can write $\sin^5 x$ as $(\sin^2 x)^2 \sin x = (1 - \cos^2 x)^2 \sin x$. Substituting these into the integral: $I = \int (1 - t^2)^2 t^4 (-dt) = - \int (1 - 2t^2 + t^4) t^4 \, dt$ $I = - \int (t^4 - 2t^6 + t^8) \, dt$ $I = - (\frac{t^5}{5} - \frac{2t^7}{7} + \frac{t^9}{9}) + c$ $I = - \frac{t^5}{5} + \frac{2t^7}{7} - \frac{t^9}{9} + c$ Substituting $t = \cos x$ back: $I = - \frac{1}{5}\cos^5 x + \frac{2}{7}\cos^7 x - \frac{1}{9}\cos^9 x + c$.

$\int \sin^5 x \cos^4 x \, dx = $

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