How do you evaluate the integral int sin^3xsqrt(1-cosx)?

Jan 20, 2017

$\int {\sin}^{3} x \sqrt{1 - \cos x} \mathrm{dx} = \frac{2}{7} {\left(1 - \cos x\right)}^{\frac{5}{2}} \left(\frac{9}{5} + \cos x\right) + C$

Explanation:

Write:

${\sin}^{3} x = \sin x \cdot {\sin}^{2} x = \sin x \left(1 - {\cos}^{2} x\right) = \sin x \left(1 - \cos x\right) \left(1 + \cos x\right)$

so that:

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

substitute:

$t = 1 - \cos x$
$\mathrm{dt} = \sin x \mathrm{dx}$

$\int {\sin}^{3} x \sqrt{1 - \cos x} \mathrm{dx} = \int {t}^{\frac{3}{2}} \left(2 - t\right) \mathrm{dt} = 2 \int {t}^{\frac{3}{2}} \mathrm{dt} - \int {t}^{\frac{5}{2}} \mathrm{dt} = \frac{4}{5} {t}^{\frac{5}{2}} - \frac{2}{7} {t}^{\frac{7}{2}} + C = \frac{2}{7} {t}^{\frac{5}{2}} \left(\frac{14}{5} - t\right) + C$

Substituting back:

$\int {\sin}^{3} x \sqrt{1 - \cos x} \mathrm{dx} = \frac{2}{7} {\left(1 - \cos x\right)}^{\frac{5}{2}} \left(\frac{9}{5} + \cos x\right) + C$