# How do you integrate int x^2/sqrt(25-x^2) using trig substitutions?

Jul 31, 2016

$\frac{25}{2} {\sin}^{- 1} \left(\frac{x}{5}\right) - \frac{x}{2} \sqrt{25 - {x}^{2}} + C$

#### Explanation:

Let $x = 5 \sin \left(u\right) \implies \mathrm{dx} = 5 \cos \left(u\right) \mathrm{du}$

Integral becomes

$\int \frac{25 {\sin}^{2} \left(u\right) 5 \cos \left(u\right)}{\sqrt{25 - 25 {\sin}^{2} \left(u\right)}} \mathrm{du}$

$75 \int \frac{{\sin}^{2} \left(u\right) \cos \left(u\right)}{5 \sqrt{1 - {\sin}^{2} \left(u\right)}} \mathrm{du}$

Recall that ${\sin}^{2} x + {\cos}^{2} x = 1 \implies {\cos}^{2} x = 1 - {\sin}^{2} x$

$25 \int \frac{{\sin}^{2} \left(u\right) \cos \left(u\right)}{\sqrt{{\cos}^{2} \left(u\right)}} \mathrm{du} = 25 \int {\sin}^{2} \left(u\right) \mathrm{du}$

Given $\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \left(\theta\right)$

We have that ${\sin}^{2} \left(u\right) = \frac{1}{2} \left(1 - \cos \left(2 u\right)\right)$

Integral becomes

$\frac{25}{2} \int 1 - \cos \left(2 u\right) \mathrm{du} = \frac{25}{2} \left[u - \frac{1}{2} \sin \left(2 u\right)\right] + C$

$x = 5 \sin \left(u\right) \implies u = {\sin}^{- 1} \left(\frac{x}{5}\right)$

$\frac{25}{2} \left[{\sin}^{- 1} \left(\frac{x}{5}\right) - \frac{1}{2} \sin \left(2 {\sin}^{- 1} \left(\frac{x}{5}\right)\right)\right] + C$

As is, I don't really want to deal with that second term, so what we do is go back a step and expand $\sin \left(2 u\right) = 2 \sin \left(u\right) \cos \left(u\right)$

$\frac{25}{2} \left[{\sin}^{- 1} \left(\frac{x}{5}\right) - \sin \left({\sin}^{- 1} \left(\frac{x}{5}\right)\right) \cos \left({\sin}^{- 1} \left(\frac{x}{5}\right)\right)\right] + C$

Obviously sin and arcsin will cancel out but how do we deal with cos and arcsin? My preferred method is using a simple right angled triangle.

Consider $y = {\sin}^{- 1} \left(x\right)$

$\implies x = \sin \left(y\right)$

Given sine is defined as opposite/hypotenuse this means that in our right angled triangle we have opposite side to the angle $y$ of length $x$ and hypotenuse of length 1. By Pythagoras the adjacent side will therefore be $\sqrt{1 - {x}^{2}}$.

$\cos \left(y\right) = \cos \left({\sin}^{- 1} \left(x\right)\right) = \left(\text{Adjacent")/("Hypotenuse}\right) = \sqrt{1 - {x}^{2}}$

So $\cos \left({\sin}^{- 1} \left(\frac{x}{5}\right)\right) = \sqrt{1 - {\left(\frac{x}{5}\right)}^{2}} = \frac{1}{5} \sqrt{25 - {x}^{2}}$

$\frac{25}{2} \left[{\sin}^{- 1} \left(\frac{x}{5}\right) - \left(\frac{x}{5}\right) \left(\frac{1}{5} \sqrt{25 - {x}^{2}}\right)\right] + C$
$\frac{25}{2} \left[{\sin}^{- 1} \left(\frac{x}{5}\right) - \frac{x}{25} \sqrt{25 - {x}^{2}}\right] + C$
$\frac{25}{2} {\sin}^{- 1} \left(\frac{x}{5}\right) - \frac{x}{2} \sqrt{25 - {x}^{2}} + C$