# How do you integrate int 1/(x^2sqrt(4-x^2)) by trigonometric substitution?

Jul 26, 2016

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{4 - {x}^{2}}} = - \frac{\sqrt{4 - {x}^{2}}}{4 x} + C$

#### Explanation:

Let $x = 2 \sin \left(u\right) \implies \mathrm{dx} = 2 \cos \left(u\right) \mathrm{du}$

$\int \frac{\mathrm{dx}}{{x}^{2} \sqrt{4 - {x}^{2}}} = \int \frac{2 \cos \left(u\right)}{4 {\sin}^{2} \left(u\right) \sqrt{4 - 4 {\sin}^{2} \left(u\right)}} \mathrm{du}$

$= \int \frac{2 \cos \left(u\right)}{8 {\sin}^{2} \left(u\right) \sqrt{1 - {\sin}^{2} \left(u\right)}} \mathrm{du}$

$= \int \frac{2 \cos \left(u\right)}{8 {\sin}^{2} \left(u\right) \sqrt{{\cos}^{2} \left(u\right)}} \mathrm{du}$

$= \int \frac{2 \cos \left(u\right)}{8 {\sin}^{2} \left(u\right) \cos \left(u\right)} \mathrm{du}$

$= \frac{1}{4} \int \frac{\mathrm{du}}{{\sin}^{2} \left(u\right)} = \frac{1}{4} \int {\csc}^{2} \left(u\right) \mathrm{du}$

You can look up that the integral of ${\csc}^{2} \left(x\right)$ is equal to $- \cot \left(x\right)$ but we shall derive this anyway.

$\int {\csc}^{2} \left(x\right) \mathrm{dx} = \int \frac{\mathrm{dx}}{\sin} ^ 2 \left(x\right)$

Divide top and bottom by ${\cos}^{2} \left(x\right)$

$\int \frac{{\sec}^{2} \left(x\right)}{{\tan}^{2} \left(x\right)} \mathrm{dx}$

Let $z = \tan \left(x\right) \implies \mathrm{dz} = {\sec}^{2} \left(x\right) \mathrm{dx}$

$\int \frac{\mathrm{dz}}{{z}^{2}} = - \frac{1}{z} = - \frac{1}{\tan \left(x\right)} = - \cot \left(x\right)$

Hence:

$\frac{1}{4} \int {\csc}^{2} \left(u\right) \mathrm{du} = - \frac{1}{4} \cot \left(u\right) + C$

$u = {\sin}^{- 1} \left(\frac{x}{2}\right)$

$\therefore I = - \frac{1}{4} \cot \left({\sin}^{- 1} \left(\frac{x}{2}\right)\right) + C$

This can look a little scary but it's actually quite simple, we just need to draw a triangle.

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

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

Sine is Opposite/Hypotenuse so we have triangle with opposite side $a$ and Hypotenuse equal to $1$. Can work out the other side with pythagoras' to be $\sqrt{1 - {a}^{2}}$.

Now, the tangent function is opposite/adjacent, hence:

$\tan \left(y\right) = \frac{a}{\sqrt{1 - {a}^{2}}}$

$\cot \theta = \frac{1}{\tan \theta} \implies \cot \left(y\right) = \frac{\sqrt{1 - {a}^{2}}}{a}$

So $- \frac{1}{4} \cot \left({\sin}^{- 1} \left(\frac{x}{2}\right)\right) = - \frac{1}{4} \frac{\sqrt{1 - {\left(\frac{x}{2}\right)}^{2}}}{\frac{x}{2}}$

$= - \frac{1}{2} \frac{\sqrt{1 - \left({x}^{2} / 4\right)}}{x} = - \frac{1}{4} \frac{\sqrt{4 - {x}^{2}}}{x} = - \frac{\sqrt{4 - {x}^{2}}}{4 x}$