# Question c645a

Jan 22, 2018

$\int {\cos}^{2} x {\sin}^{5} x \mathrm{dx} = - {\cos}^{3} \frac{x}{3} + \frac{2 {\cos}^{5} x}{5} - {\cos}^{7} \frac{x}{7} + C$

#### Explanation:

Given: $\int {\cos}^{2} x {\sin}^{5} x \mathrm{dx}$

Let u=cosx=>color(blue)(du=-sinxdx

We can strip a $\sin x$ out of the integral and rewrite the integral as

color(blue)(-)intcos^2xsin^4xcolor(blue)(sinxdx

Simplify

color(blue)(-)intcos^2x(sin^2x)^2color(blue)(sinxdx

Note: I will not make the substitution right away, I will still have to manipulate the integral to make it easier to solve. With that being said, whatever is in color(blue)("blue" will not be involved in the following procedure until it's time to make the subsitution.

Convert the sines to cosines

Use the identity ${\cos}^{2} x + {\sin}^{2} x = 1 \implies {\sin}^{2} x = 1 - {\cos}^{2} x$

color(blue)(-)intcos^2x(1-cos^2x)^2color(blue)(sinxdx

Expand ${\left(1 - {\cos}^{2} x\right)}^{2}$

color(blue)(-)intcos^2x(1-2cos^2x+cos^4x)color(blue)(sinxdx

Distribute ${\cos}^{2} x$ to each term

color(blue)(-)intcos^2x-2cos^4x+cos^6xcolor(blue)(sinxdx#

Now I will make use of the substitution:

$- \int {u}^{2} - 2 {u}^{4} + {u}^{6} \mathrm{du}$

Integrating each term we get...

$= - \left[{u}^{3} / 3 - \frac{2 {u}^{5}}{5} + {u}^{7} / 7\right] + C$

$= - {u}^{3} / 3 + \frac{2 {u}^{5}}{5} - {u}^{7} / 7 + C$

Reverse the substiution:

$= - {\cos}^{3} \frac{x}{3} + \frac{2 {\cos}^{5} x}{5} - {\cos}^{7} \frac{x}{7} + C$