# How do you integrate int sqrt(13+25x^2) using trig substitutions?

Nov 25, 2016

To integrate an irrational function with a sum of squares under the root, you use the substitution with the tangent.

#### Explanation:

First reduce the radical to a sum of squares:

$\int \left(\sqrt{13 + 25 {x}^{2}}\right) \mathrm{dx} = \frac{1}{5} \int \left(\sqrt{{\left(\frac{\sqrt{13}}{5}\right)}^{2} + {x}^{2}}\right) \mathrm{dx}$

Pose $x = \frac{\sqrt{13}}{5} \tan \left(t\right)$ and as $\frac{d \tan \left(t\right)}{\mathrm{dt}} = {\sec}^{2} \left(t\right)$

$\mathrm{dx} = \frac{\sqrt{13}}{5} {\sec}^{2} \left(t\right) \mathrm{dt}$

int(sqrt(13+25x^2))dx = 1/5int(sqrt(13/25+13/25tan^2(t))sqrt(13)/5sec^2(t)dt = 13/125int(sqrt(1+tan^2(t))sec^2(t)dt

Use the trigonometric identity:

$1 + {\tan}^{2} t = {\sec}^{2} t$

int(sqrt(13+25x^2))dx = 13/125int(sqrt(sec^2(t))sec^2(t)dt =
$= \frac{13}{125} \int | \sec \left(t\right) {|}^{3} \mathrm{dt}$.

Now, to integrate this last integral let's limit ourselves in the interval $t \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ where $\sec \left(t\right)$ is positive and use integration by parts:

$\int {\sec}^{3} \left(t\right) \mathrm{dt} = \int \left(\sec t \cdot {\sec}^{2} t\right) \mathrm{dt} = \int \sec t \cdot d \left(\tan t\right) = \sec t \tan t - \int \tan t \cdot d \left(\sec t\right) = \sec t \tan t - \int {\tan}^{2} t \cdot \sec t \mathrm{dt} =$

but ${\tan}^{2} t = {\sec}^{2} t - 1$, so

$\int {\sec}^{3} \left(t\right) \mathrm{dt} = \sec t \tan t + \int \left(\sec t - {\sec}^{3} t\right) \mathrm{dt} = \sec t \tan t + \int \sec t \mathrm{dt} - \int {\sec}^{3} \left(t\right) t \mathrm{dt}$

or

$2 \int {\sec}^{3} \left(t\right) \mathrm{dt} = \sec t \tan t + \int \sec t \mathrm{dt}$

Finally:

$\int {\sec}^{3} \left(t\right) \mathrm{dt} = \frac{1}{2} \sec t \tan t + \frac{1}{2} \ln \left(| \sec t + \tan t |\right)$

Now reverse the substitution:

$\tan t = \frac{5}{\sqrt{13}} x$

and $\sec t = \sqrt{1 + {\tan}^{2} \left(t\right)} = \sqrt{1 + \frac{25}{13} {x}^{2}}$

So:

$\int \left(\sqrt{13 + 25 {x}^{2}}\right) \mathrm{dx} = \frac{5}{2 \sqrt{13}} x \sqrt{1 + \frac{25}{13} {x}^{2}} + \frac{1}{2} \ln \left(| \frac{5}{\sqrt{13}} x + \sqrt{1 + \frac{25}{13} {x}^{2}} |\right)$