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

Dec 22, 2016

$\sqrt{{x}^{2} + 1} + C$

#### Explanation:

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

$\implies \int \tan \frac{\theta}{\sqrt{\left({\tan}^{2} \theta + 1\right)}} {\sec}^{2} \theta d \theta$

$\implies \int \tan \frac{\theta}{\sqrt{{\sec}^{2} \theta}} {\sec}^{2} \theta d \theta$

$\implies \int \tan \frac{\theta}{\sec} \theta {\sec}^{2} \theta d \theta$

$\implies \int \tan \theta \sec \theta d \theta$

This is a common integral--$\int \left(\tan x \sec x\right) \mathrm{dx} = \sec x + C$.

$\implies \sec \theta + C$

We now draw an imaginary triangle.

The definition of $\sec \theta$ is "hypotenuse"/("side adjacent" theta) because $\sec x = \frac{1}{\cos} x$. In this image, $\sec \theta = \sqrt{{x}^{2} + 1}$.

Therefore, the integral can be simplified to $\sqrt{{x}^{2} + 1} + C$.

Hopefully this helps!

Dec 23, 2016

By inspection rather than trig substitution.

#### Explanation:

Notice that the $x$ in the top is proportional to the derivative of the function under the square root. So just write down ${\left({x}^{2} + 1\right)}^{- \frac{1}{2} + 1}$ as a trial (rather like integrating any power of $x$) and differentiate it. By the chain rule, the function under the square root will provide the $2 x$ and the $2$'s will cancel, so the trial was an immediate success.

Alternatively substitute $u = {x}^{2} + 1$, $\mathrm{dx} = \frac{\mathrm{du}}{2 \sqrt{u - 1}}$ and the integral becomes $\left(\frac{1}{2}\right) \int {u}^{- \frac{1}{2}} \mathrm{du}$ = ${u}^{\frac{1}{2}} + C$=$\sqrt{{x}^{2} + 1} + C$