# How do you integrate int sqrt(1+x^2)/xdx using trigonometric substitution?

Mar 10, 2018

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

#### Explanation:

We let $x = \tan \theta$. Then $\mathrm{dx} = {\sec}^{2} \theta d \theta$.

$I = \int \frac{\sqrt{1 + {\tan}^{2} \theta}}{\tan} \theta \cdot {\sec}^{2} \theta d \theta$

$I = \int \frac{\sqrt{{\sec}^{2} \theta}}{\tan} \theta \cdot {\sec}^{2} \theta d \theta$

$I = \int {\sec}^{3} \frac{\theta}{\tan} \theta d \theta$

$I = \int \frac{\frac{1}{\cos} ^ 3 \theta}{\sin \frac{\theta}{\cos} \theta} d \theta$

$I = \int \csc \theta {\sec}^{2} \theta$

$I = \int \csc \theta \left(1 + {\tan}^{2} \theta\right) d \theta$

$I = \int \csc \theta + \csc \theta {\tan}^{2} \theta d \theta$

$I = \int \csc \theta + \sec \theta \cos \theta d \theta$

Now these are two known integrals.

$I = \sec \theta - \ln | \csc \theta + \cot \theta | + C$

IF $\frac{x}{1} = \tan \theta$, then $\sec \theta = \sqrt{{x}^{2} + 1}$ and $\csc \theta = \frac{\sqrt{{x}^{2} + 1}}{x}$ and $\cot \theta = \frac{1}{x}$.

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

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

Hopefully this helps!