int cos^5 x , dx = | vedclass

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Question

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

Vedclass · Accepted Answer

$\sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + c$

Vedclass · Answer

(A) We have $I = \int \cos^5 x \, dx$. We can write this as $I = \int \cos^4 x \cdot \cos x \, dx = \int (1 - \sin^2 x)^2 \cos x \, dx$. Let $t = \sin x$,then $dt = \cos x \, dx$. Substituting these into the integral,we get $I = \int (1 - t^2)^2 \, dt$. Expanding the integrand,$I = \int (1 - 2t^2 + t^4) \, dt$. Integrating term by term,$I = t - \frac{2t^3}{3} + \frac{t^5}{5} + c$. Substituting $t = \sin x$ back,we get $I = \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + c$.

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

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