# How do you integrate int 1/sqrt(7-3x^2) by trigonometric substitution?

Oct 30, 2016

$\int \frac{1}{\sqrt{7 - 3 {x}^{2}}} \mathrm{dx} = \left(\frac{\sqrt{3}}{3}\right) {\sin}^{-} 1 \left(\sqrt{\frac{3}{7}} x\right) + C$

#### Explanation:

Given:

$\int \frac{1}{\sqrt{7 - 3 {x}^{2}}} \mathrm{dx}$

Divide by $\sqrt{3}$ so that the integrand is in the form $\frac{1}{\sqrt{{a}^{2} - {x}^{2}}}$

$\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{\frac{7}{3} - {x}^{2}}} \mathrm{dx}$

Please notice that ${a}^{2} = \frac{7}{3}$, therefore, $a = \sqrt{\frac{7}{3}}$

The reference for Trignometric substitutions says to substitute $x = a \sin \left(t\right)$

let $x = \sqrt{\frac{7}{3}} \sin \left(t\right)$, then $\mathrm{dx} = \sqrt{\frac{7}{3}} \cos \left(t\right) \mathrm{dt}$

$\frac{1}{\sqrt{3}} \int \frac{\sqrt{\frac{7}{3}} \cos \left(t\right)}{\sqrt{\frac{7}{3} - \frac{7}{3} {\sin}^{2} \left(t\right)}} \mathrm{dt}$

From the same reference, it says to substitute ${a}^{2} {\cos}^{2} \left(t\right)$ for ${a}^{2} - {a}^{2} {\sin}^{2} \left(t\right)$:

$\frac{1}{\sqrt{3}} \int \frac{\sqrt{\frac{7}{3}} \cos \left(t\right)}{\sqrt{\frac{7}{3} {\cos}^{2} \left(t\right)}} \mathrm{dt} =$

$\frac{1}{\sqrt{3}} \int \frac{\sqrt{\frac{7}{3}} \cos \left(t\right)}{\sqrt{\frac{7}{3}} \cos \left(t\right)} \mathrm{dt} =$

$\frac{1}{\sqrt{3}} \int \mathrm{dt} = \frac{\sqrt{3} \left(t\right)}{3} + C$

Solve the original substitution for t:

$x = \sqrt{\frac{7}{3}} \sin \left(t\right)$

$\sin \left(t\right) = \sqrt{\frac{3}{7}} x$

$t = {\sin}^{-} 1 \left(\sqrt{\frac{3}{7}} x\right)$

Substitute ${\sin}^{-} 1 \left(\sqrt{\frac{3}{7}} x\right)$ for $t$

$\int \frac{1}{\sqrt{7 - 3 {x}^{2}}} \mathrm{dx} = \frac{\sqrt{3} \left({\sin}^{-} 1 \left(\sqrt{\frac{3}{7}} x\right)\right)}{3} + C$