# How do you integrate int x/sqrt(144+x^2)dx using trigonometric substitution?

Mar 4, 2016

$\int \frac{x}{\sqrt{144 + {x}^{2}}} \mathrm{dx} = \sqrt{144 + {x}^{2}} + C$
Refer below for explanation.

#### Explanation:

Step 1: Draw It!
The first thing to do with trig substitution problems, especially if you have time, is to draw them out. Note that the expression in the denominator - $\sqrt{144 + {x}^{2}}$ - resembles the Pythagorean Theorem, which says for any right triangle, the hypotenuse is $\sqrt{{a}^{2} + {b}^{2}}$. We could rewrite $\sqrt{144 + {x}^{2}}$ as $\sqrt{{\left(12\right)}^{2} + {x}^{2}}$, which looks pretty much the same as the theorem, with $a = 12$ and $b = x$. Using this info, we can make a pretty picture of this problem:

Step 2: Define a Few Things
From the image, we see that $\tan \left(\theta\right) = \frac{x}{12}$ and furthermore, $12 \tan \left(\theta\right) = x$. Taking the derivative of both sides, we see that $\frac{\mathrm{dx}}{d \theta} = 12 {\sec}^{2} \left(\theta\right)$. Multiplying both sides by $d \theta$ yields $\mathrm{dx} = 12 {\sec}^{2} \left(\theta\right) d \theta$.

Step 3: Trigonometric Substitution
Now we can finally take this information and apply it to the problem. Making substitutions, our integral becomes:
$\int \frac{12 \tan \left(\theta\right)}{\sqrt{144 + {\left(12 \tan \left(\theta\right)\right)}^{2}}} 12 {\sec}^{2} \left(\theta\right) d \theta$

Step 4: Simplification
This tends to be pretty long, so bear with me here.
$= 144 \int \frac{\tan \left(\theta\right) {\sec}^{2} \left(\theta\right)}{\sqrt{144 + 144 {\tan}^{2} \left(\theta\right)}} d \theta$
$= 144 \int \frac{\tan \left(\theta\right) {\sec}^{2} \left(\theta\right)}{\sqrt{144 \left(1 + {\tan}^{2} \left(\theta\right)\right)}} d \theta$
$= 144 \int \frac{\tan \left(\theta\right) {\sec}^{2} \left(\theta\right)}{12 \sqrt{1 + {\tan}^{2} \left(\theta\right)}} d \theta$
$= 12 \int \frac{\tan \left(\theta\right) {\sec}^{2} \left(\theta\right)}{\sqrt{1 + {\tan}^{2} \left(\theta\right)}} d \theta$

Remember the Pythagorean Identities from trig? One of those identities is $1 + {\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right)$. Using that fact,
$= 12 \int \frac{\tan \left(\theta\right) {\sec}^{2} \left(\theta\right)}{\sqrt{{\sec}^{2} \left(\theta\right)}} d \theta$
$= 12 \int \frac{\tan \left(\theta\right) {\sec}^{2} \left(\theta\right)}{\sec \left(\theta\right)} d \theta$
$= 12 \int \tan \left(\theta\right) \sec \left(\theta\right) d \theta$

Hm...think about it for a moment. What function's derivative is $\tan \left(\theta\right) \sec \left(\theta\right)$? Secant, of course! Which means our integral is...
$12 \sec \left(\theta\right) + C$.

You might think we're done, but wait a minute. The problem was given to us in terms of $x$; our answer is in terms of $\theta$. That leads us to...

Step 5: Reverse Substitution
Wonder why I told you to draw the problem? Because now we can clearly see from the triangle above that $\sec \left(\theta\right) = \frac{\sqrt{144 + {x}^{2}}}{12}$. Lastly, we deal with that pesky $12$ in front of secant. Note that multiplying $12$ to both sides of $\sec \left(\theta\right) = \frac{\sqrt{144 + {x}^{2}}}{12}$ gives us $12 \sec \left(\theta\right) = \sqrt{144 + {x}^{2}}$. Now that our answer is in terms of $x$, we can officially say we're done.

Final answer: $\int \frac{x}{\sqrt{144 + {x}^{2}}} \mathrm{dx} = \sqrt{144 + {x}^{2}} + C$