# How do you integrate int 1/(x^3sqrt(9-x^2))dx using trigonometric substitution?

Jan 21, 2017

$- \frac{1}{54} \left(\ln | \frac{3 + \sqrt{9 - {x}^{2}}}{x} | + \frac{3 \sqrt{9 - {x}^{2}}}{x} ^ 2\right) + C$

#### Explanation:

Whenever you have an integral with a √ of the form $\sqrt{{a}^{2} - {x}^{2}}$, use the substitution $x = a \sin \theta$.

So, let $x = 3 \sin \theta$. Then $\mathrm{dx} = 3 \cos \theta d \theta$.

$\implies \int \frac{1}{{\left(3 \sin \theta\right)}^{3} \sqrt{9 - {\left(3 \sin \theta\right)}^{2}}} \cdot 3 \cos \theta d \theta$

$\implies \int \frac{1}{27 {\sin}^{3} \theta \sqrt{9 - 9 {\sin}^{2} \theta}} \cdot 3 \cos \theta d \theta$

$\implies \int \frac{1}{27 {\sin}^{3} \theta \sqrt{9 \left(1 - {\sin}^{2} \theta\right)}} \cdot 3 \cos \theta d \theta$

$\implies \int \frac{1}{27 {\sin}^{3} \theta \sqrt{9 {\cos}^{2} \theta}} \cdot 3 \cos \theta d \theta$

$\implies \int \frac{1}{27 {\sin}^{3} \theta 3 \cos \theta} \cdot 3 \cos \theta d \theta$

$\implies \int \frac{1}{27 {\sin}^{3} \theta} d \theta$

$\implies \frac{1}{27} \int {\csc}^{3} \theta d \theta$

This can be integrated as $- \frac{\ln | \csc \theta + \cot \theta | + \cot \theta \csc \theta}{2} + C$. This is derived using integration by parts. You can see the full proof here.

$\implies - \frac{1}{54} \left(\ln | \csc \theta + \cot \theta | + \cot \theta \csc \theta\right) + C$

Now draw a triangle to determine expressions for cosecant and cotangent.

$\implies - \frac{1}{54} \left(\ln | \frac{3 + \sqrt{9 - {x}^{2}}}{x} | + \frac{3 \sqrt{9 - {x}^{2}}}{x} ^ 2\right) + C$

Hopefully this helps!