# How do you find the integral of (x^2)/(sqrt(4-(9(x^2))) ?

Jul 14, 2015

$\int {x}^{2} / \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = - \frac{1}{18} x \sqrt{4 - 9 {x}^{2}} - \frac{2}{27} {\cos}^{- 1} \left(\frac{3 x}{2}\right) + c$

#### Explanation:

For this problem to make sense $4 - 9 {x}^{2} \ge 0$, so $- \frac{2}{3} \le x \le \frac{2}{3}$. Therefore we can choose a $0 \le u \le \pi$ such that $x = \frac{2}{3} \cos u$. Using this, we can subsitute the variable x in the integral using $\mathrm{dx} = - \frac{2}{3} \sin u \mathrm{du}$: $\int {x}^{2} / \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = - \frac{4}{27} \int {\cos}^{2} \frac{u}{\sqrt{1 - {\cos}^{2} u}} \sin u \mathrm{du} = - \frac{4}{27} \int {\cos}^{2} u \mathrm{du}$ here we use that $1 - {\cos}^{2} u = {\sin}^{2} u$ and that for $0 \le u \le \pi$ $\sin u \ge 0$.

Now we use integration by parts to find $\int {\cos}^{2} u \mathrm{du} = \int \cos u \mathrm{ds} \in u = \sin u \cos u - \int \sin u \mathrm{dc} o s u = \sin u \cos u + \int {\sin}^{2} u = \sin u \cos u + \int \mathrm{du} - \int {\cos}^{2} u \mathrm{du} = \sin u \cos u + u + c - \int {\cos}^{2} u \mathrm{du}$. Therefore $\int {\cos}^{2} u \mathrm{du} = \frac{1}{2} \left(\sin u \cos u + u + c\right)$.

So we have found $\int {x}^{2} / \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = - \frac{2}{27} \left(\sin u \cos u + u + c\right)$, now we substitute $x$ back for $u$, using $u = {\cos}^{- 1} \left(\frac{3 x}{2}\right)$, so $\int {x}^{2} / \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = - \frac{1}{9} x \sin \left({\cos}^{- 1} \left(\frac{3 x}{2}\right)\right) - \frac{2}{27} {\cos}^{- 1} \left(\frac{3 x}{2}\right) + c$.

We can further simplify this by using the definition of sines and cosines in terms of triangles. For a right triangle with an angle $u$ at one of the non-right corners, $\sin u = \text{opposite side"/"longest side}$, while $\cos u = \text{adjacent side"/"longest side}$, since we know $\cos u = \frac{3 x}{2}$, we can pick the adjacent side to be $3 x$ and the longest side to be $2$. Using Pythagoras' theorem, we find the opposite side to be $\sqrt{4 - 9 {x}^{2}}$, so $\sin \left({\cos}^{- 1} \left(\frac{3 x}{2}\right)\right) = \sin u = \frac{1}{2} \sqrt{4 - 9 {x}^{2}}$. Therefore $\int {x}^{2} / \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = - \frac{1}{18} x \sqrt{4 - 9 {x}^{2}} - \frac{2}{27} {\cos}^{- 1} \left(\frac{3 x}{2}\right) + c$.