How do you find the integral of int (6sin^6x)(cos^3x)dx?

Oct 20, 2015

$I = \frac{6 {\sin}^{7} x}{7} - \frac{2 {\sin}^{9} x}{3} + C$

Explanation:

$I = \int \left(6 {\sin}^{6} x\right) \left({\cos}^{3} x\right) \mathrm{dx} = 6 \int {\sin}^{6} x {\cos}^{2} x \cos x \mathrm{dx}$

$I = 6 \int {\sin}^{6} x \left(1 - {\sin}^{2} x\right) \cos x \mathrm{dx}$

$\sin x = t \implies \cos x \mathrm{dx} = \mathrm{dt}$

$I = 6 \int {t}^{6} \left(1 - {t}^{2}\right) \mathrm{dt} = 6 \int \left({t}^{6} - {t}^{8}\right) \mathrm{dt} = 6 \left({t}^{7} / 7 - {t}^{9} / 9 + C\right)$

$I = \frac{6 {\sin}^{7} x}{7} - \frac{2 {\sin}^{9} x}{3} + C$