Integral of 1/(1+4x^2)

**Fallen186** · May 6th 2009, 12:57 PM

Problem:

$\int \frac{dx}{1+4x^2}dx$

Attempt:

$4x = tan(\theta)$
$dx = \frac{d\theta}{cos^2(\theta)}$

$\int \frac {d\theta}{(1+tan^2(\theta))*cos^2(\theta)}$

$\int \frac {d\theta}{(1 + \frac{sin^2(\theta)}{cos^2(\theta)})*cos^2(\theta) }$

$\int \frac{d\theta}{cos^2(\theta)+sin^2(\theta)}$

$\int d\theta$

$\theta + C$

My Answer:
$arctan(4x) + C$

Correct Answer:
$ \frac{1}{2} arctan(2x) + C $

**skeeter** · May 6th 2009, 01:01 PM

correction ...

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

$2x = \tan{\theta}$

**Danny** · May 6th 2009, 01:01 PM

Originally Posted by Fallen186

Problem:

$\int \frac{dx}{1+4x^2}dx$

Attempt:

$4x = tan(\theta)$
$dx = \frac{d\theta}{cos^2(\theta)}$

$\int \frac {d\theta}{(1+tan^2(\theta))*cos^2(\theta)}$

$\int \frac {d\theta}{(1 + \frac{sin^2(\theta)}{cos^2(\theta)})*cos^2(\theta) }$

$\int \frac{d\theta}{cos^2(\theta)+sin^2(\theta)}$

$\int d\theta$

$\theta + C$

My Answer:
$arctan(4x) + C$

Correct Answer:
$ \frac{1}{2} arctan(2x) + C $

Close. Try $2x = \tan \theta$