How do you integrate $\int \frac{x}{\sqrt{144 + x^{2}}} d x$ $int x/sqrt(144+x^2)dx$ using trigonometric substitution?

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How do you integrate $\int \frac{x}{\sqrt{144 + x^{2}}} d x$ $int x/sqrt(144+x^2)dx$ using trigonometric substitution?

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Answer 1 · 2016-03-04T21:20:32

$\int \frac{x}{\sqrt{144 + x^{2}}} d x = \sqrt{144 + x^{2}} + C$ $int x/sqrt(144+x^2) dx = sqrt(144+x^2)+C$
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 - $\sqrt{144 + x^{2}}$ $sqrt(144+x^2)$ - resembles the Pythagorean Theorem, which says for any right triangle, the hypotenuse is $\sqrt{a^{2} + b^{2}}$ $sqrt(a^2+b^2)$ . We could rewrite $\sqrt{144 + x^{2}}$ $sqrt(144+x^2)$ as $\sqrt{{(12)}^{2} + x^{2}}$ $sqrt((12)^2+x^2)$ , which looks pretty much the same as the theorem, with $a = 12$ $a = 12$ and $b = x$ $b = x$ . Using this info, we can make a pretty picture of this problem:
enter image source here

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

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

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

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

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

You might think we're done, but wait a minute. The problem was given to us in terms of $x$ $x$ ; our answer is in terms of $θ$ $theta$ . 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 $sec (θ) = \frac{\sqrt{144 + x^{2}}}{12}$ $sec(theta) = sqrt(144+x^2)/12$ . Lastly, we deal with that pesky $12$ $12$ in front of secant. Note that multiplying $12$ $12$ to both sides of $sec (θ) = \frac{\sqrt{144 + x^{2}}}{12}$ $sec(theta) = sqrt(144+x^2)/12$ gives us $12 sec (θ) = \sqrt{144 + x^{2}}$ $12sec(theta) = sqrt(144+x^2)$ . Now that our answer is in terms of $x$ $x$ , we can officially say we're done.

Final answer: $\int \frac{x}{\sqrt{144 + x^{2}}} d x = \sqrt{144 + x^{2}} + C$ $int x/sqrt(144+x^2) dx = sqrt(144+x^2)+C$

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How do you integrate $\int \frac{x}{\sqrt{144 + x^{2}}} d x$ $int x/sqrt(144+x^2)dx$ using trigonometric substitution?

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