How do I find the antiderivative of sin^7(x)cos^6(x)?

May 23, 2015

In this way, remembering:

$\int {\left[f \left(x\right)\right]}^{n} f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c$,

$\int {\sin}^{7} x {\cos}^{6} x \mathrm{dx} = \int \sin x {\sin}^{6} x {\cos}^{6} x \mathrm{dx} =$

$= \int \sin x {\left({\sin}^{2} x\right)}^{3} {\cos}^{6} x \mathrm{dx} = \int \sin x {\left(1 - {\cos}^{2} x\right)}^{3} {\cos}^{6} x \mathrm{dx} =$

$= \int \sin x \left(1 - 3 {\cos}^{2} x + 3 {\cos}^{4} x - {\cos}^{6} x\right) {\cos}^{6} x \mathrm{dx} =$

$= \int \left(\sin x {\cos}^{6} x - 3 \sin x {\cos}^{8} x + 3 \sin x {\cos}^{10} x - \sin x {\cos}^{12} x\right) \mathrm{dx} =$

$= - \int \left(- \sin x\right) {\cos}^{6} x \mathrm{dx} + 3 \int \left(- \sin x\right) {\cos}^{8} x \mathrm{dx} - 3 \int \left(- \sin x\right) {\cos}^{10} x \mathrm{dx} + \int \left(- \sin x\right) {\cos}^{12} x \mathrm{dx} =$

$= - {\cos}^{7} \frac{x}{7} + 3 {\cos}^{9} \frac{x}{9} - 3 {\cos}^{11} \frac{x}{11} + {\cos}^{13} \frac{x}{13} + c =$

$= - {\cos}^{7} \frac{x}{7} + {\cos}^{9} \frac{x}{3} - 3 {\cos}^{11} \frac{x}{11} + {\cos}^{13} \frac{x}{13} + c$.