# Thread: Trig Substitutions Need Help

1. ## Trig Substitutions Need Help

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

$\int \frac{(4+(4tan^2\theta)^2)^2}{x^3} dx$

$\int \frac{(4+(4tan^2\theta)^2)^2}{4tan^2\theta} d\theta$

$\int \frac{4+4tan^8\theta}{4tan^2\theta} d\theta$

$\int \frac{4}{4tan^2\theta} d\theta + \int \frac{4tan^8\theta}{4tan^2\theta} d\theta$

$\int \frac{1}{tan^2\theta} d\theta + \int tan^6\theta d\theta$

I'm stuck at here...

2. Originally Posted by circuscircus
$\int \frac{(4+x^2)^2}{x^3} dx$
Try $x = 2tan(\theta)$ instead.

-Dan

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

$\int \frac{(4+(2tan^2\theta)^2)^2}{2\tan^6\theta} dx$

$\int \frac{4+2tan^8\theta}{2tan^6\theta} dx$

$\int \frac{2+tan^8\theta}{tan^6\theta} dx$

What to do next?

4. Originally Posted by topsquark
Try $x = 2tan(\theta)$ instead.

-Dan
Originally Posted by circuscircus
$\int \frac{(4+x^2)^2}{x^3} dx$

$\int \frac{(4+(2tan^2\theta)^2)^2}{2\tan^6\theta} dx$

$\int \frac{4+2tan^8\theta}{2tan^6\theta} dx$

$\int \frac{2+tan^8\theta}{tan^6\theta} dx$

What to do next?
Not quite what I was thinking:
$\int \frac{(4+x^2)^2}{x^3} dx$

Let $x = 2tan{\theta} \implies dx = 2sec^2(\theta) d \theta$

$\int \frac{(4+x^2)^2}{x^3} dx = \int \frac{(4 + 4tan^2{\theta})^2}{8tan^3(\theta)} \cdot 2sec^2(\theta) d\theta$

$= 4 \int \frac{(1 + tan^2{\theta})^2}{tan^3(\theta)} \cdot sec^2(\theta) d\theta$

$= 4 \int \frac{sec^4{\theta}}{tan^3(\theta)} \cdot sec^2(\theta) d\theta$

$= 4 \int \frac{sec^6{\theta}}{tan^3(\theta)} d\theta$

And you can proceed from there.

However upon reflection:
$\int \frac{(4+x^2)^2}{x^3} dx = \int \frac{x^4 + 8x^2 + 16}{x^3}$

$= \int \left ( x + \frac{8}{x} + \frac{16}{x^3} \right ) dx$
is a much simpler method.

-Dan

5. Hello, circuscircus!

You seem to have missed the point of Trig Substitution . . .

$\int \frac{(4+x^2)^2}{x^3}\,dx$
Let: $x \:=\:2\tan\theta \quad\Rightarrow\quad dx \:=\:2\sec^2\!\theta\,d\theta$

Also: $4 + x^2 \:=\:4 + 4\tan^2\!\theta \:=\:4(1 + \tan^2\!\theta) \:=\:4\sec^2\!\theta$

Substitute: . $\int\frac{(4\sec^2\!\theta)^2}{(2\tan\theta)^3}\,( 2\sec^2\!\theta\,d\theta) \;=\;4\int\frac{\sec^4\!\theta}{\tan^3\!\theta}(\s ec^2\!\theta\,d\theta) \;= \;4\int\frac{(\sec^2\!\theta)^2}{\tan^3\!\theta}(\ sec^2\!\theta\,d\theta)$

. . . . $=\;4\int\frac{(\tan^2\!\theta + 1)^2}{\tan^3\!\theta}(\sec^2\!\theta\,d\theta) \;=\;4\int\frac{\tan^4\theta + 2\tan^2\theta + 1}{\tan^3\theta}(\sec^2\!\theta\,d\theta)$

. . . . $= \;4\int\left(\tan\theta + \frac{2}{\tan\theta} + (\tan\theta)^{-3}\right)(\sec^2\!\theta\,d\theta)$

Let: $u \,=\,\tan\theta\quad\Rightarrow\quad du \,=\,\sec^2\!\theta\,d\theta$

Substitute: . $4\int\left(u + \frac{2}{u} + u^{-3}\right)\,du$

Can you finish it now?

6. $
\int\frac{(4\sec^2\!\theta)^2}{(2\tan\theta)^3}\,( 2\sec^2\!\theta\,d\theta) \;=\;4\int\frac{\sec^4\!\theta}{\tan^3\!\theta}(\s ec^2\!\theta\,d\theta) \;= \;4\int\frac{(\sec^2\!\theta)^2}{\tan^3\!\theta}(\ sec^2\!\theta\,d\theta)
$

Ok gotcha, but I'm confused why (4sec0^2)^2 / (2tan0)^3 is 4... shouldn't it be 4^2 = 16 and 2^3=8, so 16/8 = 2?