# How do you integrate int sec^2x/(4-tan^2x)^(3/2) by trigonometric substitution?

Sep 11, 2016

$\tan \frac{x}{4 \sqrt{4 - {\tan}^{2} x}} + C$

#### Explanation:

We have:

$I = \int {\sec}^{2} \frac{x}{4 - {\tan}^{2} x} ^ \left(\frac{3}{2}\right) \mathrm{dx}$

We will first use the non-trigonometric substitution $u = \tan x$, implying that $\mathrm{du} = {\sec}^{2} x \mathrm{dx}$:

$I = \int \frac{\mathrm{du}}{4 - {u}^{2}} ^ \left(\frac{3}{2}\right)$

Now we will apply the trigonometric substitution $u = 2 \sin \theta$. Recall that this implies that $\mathrm{du} = 2 \cos \theta d \theta$.

$I = \int \frac{2 \cos \theta d \theta}{4 - 4 {\sin}^{2} \theta} ^ \left(\frac{3}{2}\right)$

$= \int \frac{2 \cos \theta d \theta}{{4}^{\frac{3}{2}} {\left(1 - {\sin}^{2} \theta\right)}^{\frac{3}{2}}}$

Note that $1 - {\sin}^{2} \theta = {\cos}^{2} \theta$:

$I = \int \frac{2 \cos \theta d \theta}{8 {\left({\cos}^{2} \theta\right)}^{\frac{3}{2}}}$

$= \int \frac{\cos \theta d \theta}{4 {\cos}^{3} \theta}$

$= \frac{1}{4} \int \frac{d \theta}{\cos} ^ 2 \theta$

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

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

Recall that $u = 2 \sin \theta$, so $\theta = \arcsin \left(\frac{u}{2}\right)$.

$I = \frac{1}{4} \tan \left(\arcsin \left(\frac{u}{2}\right)\right) + C$

We can simplify this: draw the triangle where sine is $\frac{u}{2}$, that is, the opposite side is $u$ and the hypotenuse is $2$. Through the Pythagorean theorem, we see that the adjacent side is $\sqrt{4 - {u}^{2}}$.

Thus, the tangent is opposite over adjacent, or:

$I = \frac{1}{4} \left(\frac{u}{\sqrt{4 - {u}^{2}}}\right) + C$

Now, since $u = \tan x$:

$I = \tan \frac{x}{4 \sqrt{4 - {\tan}^{2} x}} + C$