# How do you integrate int dx/sqrt(16+x^2)^2 by trigonometric substitution?

Nov 9, 2016

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Assuming you meant: $\int \frac{\mathrm{dx}}{\sqrt{16 + {x}^{2}}}$

Let $x = 4 \tan \theta$. This implies that $\mathrm{dx} = 4 {\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta$. Substituting these in yields:

$= \int \frac{4 {\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta}{\sqrt{16 + 16 {\tan}^{2} \theta}} = \frac{4}{\sqrt{16}} \int {\sec}^{2} \frac{\theta}{\sqrt{1 + {\tan}^{2} \theta}} d \theta$

Recall that $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$:

$= \int {\sec}^{2} \frac{\theta}{\sec} \theta d \theta = \int \sec \theta \textcolor{w h i t e}{.} d \theta = \ln \left\mid \tan \theta + \sec \theta \right\mid + C$

Rewriting in terms of tangent since our substitution is $\tan \theta = \frac{x}{4}$:

$= \ln \left\mid \tan \theta + \sqrt{1 + {\tan}^{2} \theta} \right\mid + C = \ln \left\mid \frac{x}{4} + \sqrt{1 + {x}^{2} / 16} \right\mid + C$

$= \ln \left\mid \frac{x}{4} + \frac{1}{4} \sqrt{16 + {x}^{2}} \right\mid + C$

Factoring the $\frac{1}{4}$ and moving it out of the integral as the constant $\ln \left(\frac{1}{4}\right)$ through the $\log \left(A B\right) = \log \left(A\right) + \log \left(B\right)$ rule, we see it combines with the constant of integration:

$= \ln \left\mid x + \sqrt{16 + {x}^{2}} \right\mid + C$

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Assuming you meant: $\int \frac{\mathrm{dx}}{\sqrt{16 + {x}^{2}}} ^ 2$

The square root and the exponent cancel, leaving just:

$= \int \frac{\mathrm{dx}}{16 + {x}^{2}}$

Now use the same substitution as before, $x = 4 \tan \theta$ such that $\mathrm{dx} = 4 {\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta$.

$= \int \frac{4 {\sec}^{2} \theta \textcolor{w h i t e}{.} d \theta}{16 + 16 {\tan}^{2} \theta} = \frac{1}{4} \int {\sec}^{2} \frac{\theta}{1 + {\tan}^{2} \theta} d \theta = \frac{1}{4} \int d \theta$

$= \frac{1}{4} \theta + C$

From the substitution $x = 4 \tan \theta$ solving for $\theta$ yields:

$= \frac{1}{4} \arctan \left(\frac{x}{4}\right) + C$