# How do you integrate int 1/(usqrt(5-u^2)) by trigonometric substitution?

Jun 15, 2017

The answer is $- \frac{1}{\sqrt{5}} \ln | \frac{\sqrt{5} + \sqrt{5 - {u}^{2}}}{u} | + C$

#### Explanation:

Use the substitution $u = \sqrt{5} \sin \theta$. Then $\mathrm{du} = \sqrt{5} \cos \theta d \theta$. Calling the integral $I$, we have:

$I = \int \frac{1}{\sqrt{5} \sin \theta \sqrt{5 - {\left(\sqrt{5} \sin \theta\right)}^{2}}} \cdot \sqrt{5} \cos \theta d \theta$

$I = \int \frac{1}{\sqrt{5} \sin \theta \sqrt{5 - 5 {\sin}^{2} \theta}} \cdot \sqrt{5} \cos \theta d \theta$

We can now simplify the square root in the denominator.

Since: $\sqrt{5 - 5 {\sin}^{2} \theta} = \sqrt{5 \left(1 - {\sin}^{2} \theta\right)} = \sqrt{5 \left({\cos}^{2} \theta\right)} = \sqrt{5} \cos \theta$ because ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$.

$I = \int \frac{1}{\sqrt{5} \sin \theta \sqrt{5} \cos \theta} \cdot \sqrt{5} \cos \theta d \theta$

$I = \int \frac{1}{\sqrt{5} \sin \theta} d \theta$

$I = \int \frac{1}{\sqrt{5}} \csc \theta d \theta$

The integral of cosecant is known.

$\textcolor{b l u e}{\int \csc x \mathrm{dx} = - \ln | \csc x + \cot x |}$

Therefore:

$I = - \frac{1}{\sqrt{5}} \ln | \csc \theta + \cot \theta |$

From our initial substitution, we know that $\sin \theta = \frac{u}{\sqrt{5}}$ and therefore $\csc \theta = \frac{1}{\sin} \theta = \frac{\sqrt{5}}{u}$. Also, $\cot \theta = \frac{\sqrt{5 - {u}^{2}}}{u}$.

$I = - \frac{1}{\sqrt{5}} \ln | \frac{\sqrt{5}}{u} + \frac{\sqrt{5 - {u}^{2}}}{u} |$

Simplify, and don't forget to add the constant of integration at the end!

$I = - \frac{1}{\sqrt{5}} \ln | \frac{\sqrt{5} + \sqrt{5 - {u}^{2}}}{u} | + C$

Hopefully this helps!