How do you evaluate the integral int cos^3xsqrtsinx?

Jan 21, 2017

$\frac{2}{3} {\left(\sin x\right)}^{\frac{3}{2}} - \frac{2}{7} {\left(\sin x\right)}^{\frac{7}{2}} + C$

Explanation:

Let $u = \sin x$.

Then $\mathrm{du} = \cos x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{\cos} x$

$\implies \int {\cos}^{3} x \sqrt{u} \cdot \frac{\mathrm{du}}{\cos} x$

$\implies \int {\cos}^{2} x \sqrt{u} \mathrm{du}$

We can rewrite ${\cos}^{2} x$ as $1 - {\sin}^{2} x$.

$\implies \int \left(1 - {\sin}^{2} x\right) \sqrt{u} \mathrm{du}$

Since $u = \sin x$, ${u}^{2} = {\sin}^{2} x$

$\implies \int \left(1 - {u}^{2}\right) \sqrt{u} \mathrm{du}$

$\implies \int \left(1 - {u}^{2}\right) {u}^{\frac{1}{2}} \mathrm{du}$

$\implies \int {u}^{\frac{1}{2}} - {u}^{\frac{5}{2}} \mathrm{du}$

This can be evaluate as $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$

$\implies \frac{2}{3} {u}^{\frac{3}{2}} - \frac{2}{7} {u}^{\frac{7}{2}} + C$

$\implies \frac{2}{3} {\left(\sin x\right)}^{\frac{3}{2}} - \frac{2}{7} {\left(\sin x\right)}^{\frac{7}{2}} + C$

Hopefully this helps!