# How do you integrate int 1/sqrt(9x^2-18x+25)  using trigonometric substitution?

Feb 27, 2017

$\frac{1}{3} \ln | \frac{3 \left(x - 1\right) + \sqrt{9 {\left(x - 1\right)}^{2} + 16}}{4} | + C$

#### Explanation:

Your first step will be to complete the square in the denominator.

$\int \frac{1}{\sqrt{9 \left({x}^{2} - 2 x\right) + 25}} \mathrm{dx}$

$\int \frac{1}{\sqrt{9 \left({x}^{2} - 2 x + 1 - 1\right) + 25}} \mathrm{dx}$

$\int \frac{1}{\sqrt{9 \left({x}^{2} - 2 x + 1\right) - 9 + 25}} \mathrm{dx}$

$\int \frac{1}{\sqrt{9 {\left(x - 1\right)}^{2} + 16}} \mathrm{dx}$

We want to convert the radical to the form $\sqrt{{a}^{2} + {x}^{2}}$ in order to make the trig substitution. We can do this by making the u-substitution $u = x - 1$. Then $\mathrm{du} = \mathrm{dx}$.

$\int \frac{1}{\sqrt{9 {u}^{2} + 16}} \mathrm{du}$

Let $u = \frac{4}{3} \tan \theta$. Then $\mathrm{du} = \frac{4}{3} {\sec}^{2} \theta d \theta$.

$\int \frac{1}{\sqrt{9 {\left(\frac{4}{3} \tan \theta\right)}^{2} + 16}} \frac{4}{3} {\sec}^{2} \theta d \theta$

$\int \frac{1}{\sqrt{16 {\tan}^{2} \theta + 16}} \cdot \frac{4}{3} {\sec}^{2} \theta d \theta$

$\int \frac{1}{\sqrt{16 \left({\tan}^{2} \theta + 1\right)}} \cdot \frac{4}{3} {\sec}^{2} \theta d \theta$

The whole point of making the trig substitution was to get rid of the radical. Here is our chance: apply the pythagorean identity $1 + {\tan}^{2} \beta = {\sec}^{2} \beta$.

$\int \frac{1}{\sqrt{16 {\sec}^{2} \theta}} \cdot \frac{4}{3} {\sec}^{2} \theta d \theta$

$\int \frac{1}{4 \sec \theta} \cdot \frac{4}{3} {\sec}^{2} \theta d \theta$

$\int \frac{1}{3} \sec \theta d \theta$

This is a known integral, whose derivation can be found here

$\frac{1}{3} \ln | \sec \theta + \tan \theta | + C$

Now comes the last part of the solution--reversing the substitutions. From our initial substitution, we know that $\frac{3 u}{4} = \tan \theta$, so the hypotenuse of the imaginary triangle would be $\sqrt{9 {u}^{2} + 16}$. Therefore, $\sec \theta = \frac{\sqrt{9 {u}^{2} + 16}}{4}$.

$\frac{1}{3} \ln | \frac{3 u}{4} + \frac{\sqrt{9 {u}^{2} + 16}}{4} | + C$

$\frac{1}{3} \ln | \frac{3 \left(x - 1\right)}{4} + \frac{\sqrt{9 {\left(x - 1\right)}^{2} + 16}}{4} | + C$

$\frac{1}{3} \ln | \frac{3 \left(x - 1\right) + \sqrt{9 {\left(x - 1\right)}^{2} + 16}}{4} | + C$

Hopefully this helps!