# Integrating: What Am I Doing Wrong?

1. Jun 19, 2012

### Hertz

1. The problem statement, all variables and given/known data

$\int \frac{dx}{(1+4x^2)^{3/2}}$

3. The attempt at a solution

$\int \frac{dx}{(1+4x^2)^{3/2}}$

Let $x = \frac{1}{4}tan(u), dx = \frac{1}{4}sec^2(u)du$

$\frac{1}{4}\int{\frac{sec^2(u)du}{(sec^2(u))^{3/2}}}$

$\frac{1}{4}\int{\frac{1}{sec(u)}}du$

$\frac{1}{4}\int{cos(u)}du$

$\frac{1}{4}sin(u) + C$

Drawing a triangle with angle u:

$tan(u) = 4x$

Therefore:

Opposite = $4x$

Adjacent = $1$

Hypotenuse = $\sqrt{1^2 + (4x)^2} = \sqrt{1 + 16x^2}$

$sin(u) = \frac{4x}{\sqrt{1 + 16x^2}}$

$\frac{1}{4}sin(u) + C$

$\frac{1}{4}(\frac{4x}{\sqrt{1 + 16x^2}}) + C$

$\frac{x}{\sqrt{1 + 16x^2}} + C$

I personally can't see what I did wrong here, but this is not the correct answer :( Both my math book and Mathematica say that the correct answer should actually be

$\frac{x}{\sqrt{1 + 4x^2}} + C$

Any help is greatly appreciated :)

Last edited by a moderator: Jun 19, 2012
2. Jun 19, 2012

### Hertz

Nevermind, I got it.

The substitution I used, $x = \frac{1}{4}tan(u)$ does not produce the result I wanted because the $\frac{1}{4}$ would be squared also. Instead, I made the substitution $x = \frac{1}{2}tan(u)$ and that led me to the correct answer.