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

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Question

(C) Let $I = \int \sin ^5 x \cos ^5 x \, dx$. We can rewrite the integral as: $I = \int \cos ^5 x \sin ^4 x \sin x \, dx = \int \cos ^5 x (1 - \cos ^2

Vedclass · Accepted Answer

$-\frac{\cos ^6 x}{60}\left(6 \sin ^4 x+3 \sin ^2 x+1\right)+c$

Vedclass · Answer

(C) Let $I = \int \sin ^5 x \cos ^5 x \, dx$. We can rewrite the integral as: $I = \int \cos ^5 x \sin ^4 x \sin x \, dx = \int \cos ^5 x (1 - \cos ^2 x)^2 \sin x \, dx$. Let $\cos x = t$,then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$. Substituting these into the integral: $I = \int t^5 (1 - t^2)^2 (-dt) = -\int t^5 (t^4 - 2t^2 + 1) \, dt$. $I = -\int (t^9 - 2t^7 + t^5) \, dt = -(\frac{t^{10}}{10} - 2 \frac{t^8}{8} + \frac{t^6}{6}) + C$. $I = -\frac{t^6}{60} (6t^4 - 15t^2 + 10) + C$. Substituting $t = \cos x$: $I = -\frac{\cos ^6 x}{60} (6 \cos ^4 x - 15 \cos ^2 x + 10) + C$. Using $\cos ^2 x = 1 - \sin ^2 x$: $I = -\frac{\cos ^6 x}{60} [6(1 - \sin ^2 x)^2 - 15(1 - \sin ^2 x) + 10] + C$. $I = -\frac{\cos ^6 x}{60} [6(1 - 2 \sin ^2 x + \sin ^4 x) - 15 + 15 \sin ^2 x + 10] + C$. $I = -\frac{\cos ^6 x}{60} [6 - 12 \sin ^2 x + 6 \sin ^4 x - 15 + 15 \sin ^2 x + 10] + C$. $I = -\frac{\cos ^6 x}{60} [6 \sin ^4 x + 3 \sin ^2 x + 1] + C$. Thus,option $(c)$ is correct.

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

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