# How do you integrate int sin^5xcos^2xdx?

Jun 21, 2017

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

#### Explanation:

I think the easiest way to get started, would be to note that ${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$

So looking at this integral, we have $\int {\sin}^{5} \left(x\right) \cdot {\cos}^{2} \left(x\right) \mathrm{dx} = \int \sin \left(x\right) \cdot {\left({\sin}^{2} \left(x\right)\right)}^{2} \cdot {\cos}^{2} \left(x\right) \mathrm{dx} = \int \sin \left(x\right) \cdot {\left(1 - {\cos}^{2} \left(x\right)\right)}^{2} \cdot {\cos}^{2} \left(x\right) \mathrm{dx}$

Why do it like this?
Because you usually want to look ahead at what is going to happen, after you have found the derivative of $u$.
In this case, using the fact that ${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$, I can rewrite and use $u '$ to get rid of either one of trigonometric functions. Had I chosen to get rid of all instances of $\cos \left(x\right)$, I would have ended up with a square root. Getting rid of $\sin \left(x\right)$ on the other hand, you can already see from now that I won't get any nasty square roots, and it looks like it's just gonne be adding together powers of $u$. That is much simpler to do.

So, let's continue :)

$u = \cos \left(x\right)$
$\frac{\mathrm{du}}{\mathrm{dx}} = - \sin \left(x\right) \iff \mathrm{dx} = \frac{\mathrm{du}}{-} \sin \left(x\right)$

$\int \frac{\sin \left(x\right) \cdot {\left(1 - {u}^{2}\right)}^{2} \cdot {u}^{2}}{-} \sin \left(x\right) \mathrm{du} = - \int {u}^{2} \cdot {\left(1 - {u}^{2}\right)}^{2} \mathrm{du}$
$= - \int {u}^{2} \cdot \left(1 + {u}^{4} - 2 {u}^{2}\right) \mathrm{du}$
$= - \int {u}^{2} + {u}^{6} - 2 {u}^{4} \mathrm{du}$
$= - \int {u}^{2} \mathrm{du} - \int {u}^{6} \mathrm{du} - \int - 2 {u}^{4} \mathrm{du}$

$= - \frac{1}{3} {u}^{3} - \frac{1}{7} {u}^{7} + \frac{2}{5} {u}^{5} + k$

$= \frac{2}{5} {\cos}^{5} \left(x\right) - \frac{1}{3} {\cos}^{3} \left(x\right) - \frac{1}{7} {\cos}^{7} \left(x\right) + k$

Here's a double check: https://www.wolframalpha.com/input/?i=derivative(2%2F5cos%5E5(x)-1%2F3cos%5E3(x)-1%2F7cos%5E7(x)%2Bk)