# How do you integrate int sqrt(4-9x^2) using trig substitutions?

Dec 30, 2016

$= \frac{1}{2} x \sqrt{4 - 9 {x}^{2}} + \frac{2}{3} {\sin}^{- 1} \left(\frac{3}{2} x\right) + C$

#### Explanation:

One obvious thing here is to use a sub to get the integrand looking like this

$\sqrt{4 - 4 {\sin}^{2} y} = 2 \cos y$, ie using the Pythagorean identity

so we can say that $9 {x}^{2} = 4 {\sin}^{2} y$, and the sub we are going to try is $x = \frac{2}{3} \sin y , \setminus \mathrm{dx} = \frac{2}{3} \cos y \setminus \mathrm{dy}$

The integration is then

$\int 2 \cos y \cdot \frac{2}{3} \cos y \setminus \mathrm{dy}$

$= \frac{4}{3} \int {\cos}^{2} y \setminus \mathrm{dy}$

use the double-angle identity $\cos 2 A = 2 {\cos}^{2} A - 1$

$= \frac{2}{3} \int \cos 2 y + 1 \setminus \mathrm{dy}$

$= \frac{2}{3} \left(\frac{1}{2} \sin 2 y + y\right) + C$

now $\sin 2 y = 2 \sin y \cos y = 2 \cdot \frac{3}{2} x \cdot \sqrt{1 - {\left(\frac{3}{2} x\right)}^{2}}$

giving

$= x \sqrt{1 - {\left(\frac{3}{2} x\right)}^{2}} + \frac{2}{3} {\sin}^{- 1} \left(\frac{3}{2} x\right) + C$

$= \frac{1}{2} x \sqrt{4 - 9 {x}^{2}} + \frac{2}{3} {\sin}^{- 1} \left(\frac{3}{2} x\right) + C$

Dec 30, 2016

$\int \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = \frac{x}{2} \sqrt{4 - 9 {x}^{2}} + \frac{2}{3} \arcsin \left(\frac{3}{2} x\right)$

#### Explanation:

Write the integral as:

$\int \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = 2 \int \sqrt{1 - {\left(\frac{3}{2} x\right)}^{2}} \mathrm{dx}$

Now substitute:

$x = \frac{2}{3} \sin t$, $\mathrm{dx} = \frac{2}{3} \cos t$

$\int \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = \frac{4}{3} \int \sqrt{1 - {\sin}^{2} t} \cos t \mathrm{dt} = \frac{4}{3} \int {\cos}^{2} t \mathrm{dt}$

To solve the last integral, we note that:

$\cos 2 x = \left(2 {\cos}^{2} x - 1\right) \implies {\cos}^{2} x = \frac{\cos 2 x + 1}{2}$

so we have:

$\int \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = \frac{2}{3} \left(\int \cos 2 t \mathrm{dt} + \int \mathrm{dt}\right) = \frac{1}{3} \left(\sin 2 t + 2 t\right)$

Substituting back x:

$\sin 2 t = 2 \sin t \cos t = 3 x \sqrt{1 - {\left(\frac{3}{2} x\right)}^{2}}$

$t = \arcsin \left(\frac{3}{2} x\right)$

Finally:

$\int \sqrt{4 - 9 {x}^{2}} \mathrm{dx} = x \sqrt{1 - {\left(\frac{3}{2} x\right)}^{2}} + \frac{2}{3} \arcsin \left(\frac{3}{2} x\right)$