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

Oct 18, 2017

$\int \setminus \frac{1}{\sqrt{2 - 5 {x}^{2}}} \setminus \mathrm{dx} = \frac{1}{\sqrt{5}} \setminus \arcsin \left(\sqrt{\frac{5}{2}} x\right) + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{\sqrt{2 - 5 {x}^{2}}} \setminus \mathrm{dx}$

Which, we can write as:

$I = \int \setminus \frac{1}{\sqrt{2 \left(1 - \frac{5}{2} {x}^{2}\right)}} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{1}{\sqrt{2} \sqrt{1 - {\left(\frac{\sqrt{5}}{\sqrt{2}} x\right)}^{2}}} \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{\sqrt{2}} \setminus \int \setminus \frac{1}{\sqrt{1 - {\left(\sqrt{\frac{5}{2}} x\right)}^{2}}} \setminus \mathrm{dx}$

We can now perform a substitution, Let:

$u = \sqrt{\frac{5}{2}} x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \sqrt{\frac{5}{2}}$
$\therefore \sqrt{\frac{2}{5}} \frac{\mathrm{du}}{\mathrm{dx}} = 1$

Substituting into the integral, we get:

$I = \frac{1}{\sqrt{2}} \setminus \int \setminus \frac{1}{\sqrt{1 - {u}^{2}}} \setminus \left(\sqrt{\frac{2}{5}}\right) \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{\sqrt{5}} \setminus \int \setminus \frac{1}{\sqrt{1 - {u}^{2}}} \setminus \mathrm{du}$

Which is a standard integral, so we have:

$I = \frac{1}{\sqrt{5}} \setminus \arcsin u + C$

Restoring the substitution:

$I = \frac{1}{\sqrt{5}} \setminus \arcsin \left(\sqrt{\frac{5}{2}} x\right) + C$