Integral of $\dfrac 1{(1+4x^2)^{3/2}}$

Question

Does anybody have any hints to solve this question?

$$\int \frac 1{(1+4x^2)^{\frac 32}}$$

I do not understand how to get $x = \frac 12 \tan(u)$ and $dx= \frac 12 \sec^2(u)$ as it says in the answer key.

Answer 1

A general guideline for trigonometric substitution is to look under the radical and see what (Pythagorean) trigonometric identity it reminds you of; here, under the radical we have

$$1 + 4x^2 = 1 + (2x)^2$$

This reminds us of the identity

$$1 + \tan^2 t = \sec^2 t$$

since we'll be able to get rid of the square root. Likewise, if we had something along the lines of

$1 - 4x^2$ we'd correspond it with $1 - \sin^2 t = \cos^2 t$ for the same reason. So we make the substitution

$$2x = \tan t \implies x = \frac 1 2 \tan t$$

Differentiating on both sides, we find (recall what the derivative of tangent is) that $$dx = \frac 1 2 \sec^2 t dt$$