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

##### 2 Answers
Dec 8, 2016

Let $x = 5 \tan \left(\theta\right)$, then dx = sec^2(theta)d"theta. To reverse the substitution, use $\theta = {\tan}^{-} 1 \left(\frac{x}{5}\right)$

#### Explanation:

$\int \frac{\mathrm{dx}}{{x}^{2} + 25}$

Let $x = 5 \tan \left(\theta\right)$, then dx = sec^2(theta)d"theta:

intsec^2(theta)/((5tan(theta))^2 + 25)d"theta =

intsec^2(theta)/(25tan^2(theta) + 25)d"theta =

1/25intsec^2(theta)/(tan^2(theta) + 1)d"theta =

Use the identity ${\sec}^{2} \left(\theta\right) = {\tan}^{2} \left(\theta\right) + 1$

1/25intsec^2(theta)/sec^2(theta)d"theta =

1/25intd"theta =

$\frac{1}{25} \theta + C =$

Reverse the substitution; substitute ${\tan}^{-} 1 \left(\frac{x}{5}\right) \text{ for } \theta$

$\frac{1}{25} {\tan}^{-} 1 \left(\frac{x}{5}\right) + C$

Dec 8, 2016

Please see below.

#### Explanation:

A good way to sort out trig substitutions is to recall the pythagorean identities in trig.

$1 - {\sin}^{2} x = {\cos}^{2} x$ is useful if we see $a - b {u}^{2}$. (We try to get $k \left(1 - {\sin}^{2} \theta\right) = k {\cos}^{2} \theta$)

In this case, we have a square plus a number. We think about how to use $t r i {g}^{2} + 1$

Eventually (perhaps by going through the list of pythagorean identities) we recall that

${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$. That gives us our substitution.

We want to have $25 {\tan}^{2} \theta + 25$, so we'll use $x = 5 {\tan}^{2} \theta$. The denominator becomes

${\left(5 \tan \theta\right)}^{2} + 25 = 25 \left({\tan}^{2} \theta + 1\right) = 25 \left({\sec}^{2} \theta\right)$

We also get $\mathrm{dx} = 5 {\sec}^{2} \theta d \theta$, so out integral becomes

$\int \frac{\mathrm{dx}}{{x}^{2} + 25} = \int \frac{5 {\sec}^{2} \theta d \theta}{25 {\sec}^{2} \theta}$

$= \frac{1}{5} \int d \theta = \frac{1}{5} \theta + C$

Since $x = 5 \tan \theta$, we have $\theta = {\tan}^{-} 1 \left(\frac{x}{5}\right) + C$

So

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