# Intgration again - Trig Sub

• Oct 31st 2007, 01:48 PM
Got5onIt
Intgration again - Trig Sub
solve:

$\int \frac{dx}{x^2 (\sqrt{x^2+1})}$

I use these trig subs:

$x=tan\theta$
$dx=sec^2 \theta \\d\theta$

after substituting and simplifying I end up here:

$\int \frac{sec\theta}{tan^2\theta} \\d\theta$

Any help in getting "unstuck" would be appreciated!
• Oct 31st 2007, 01:55 PM
Krizalid
Quote:

Originally Posted by Got5onIt
solve:

$\int \frac{dx}{x^2 (\sqrt{x^2+1})}$

Define $x=\frac1u,$ so the integral becomes to

$-\int\frac u{\sqrt{1+u^2}}\,du=-\sqrt{1+u^2}+k.$

Back substitute and you're done.
• Oct 31st 2007, 02:16 PM
Got5onIt
Thanks Kriz for the help! :)

I can't understand how $x=\frac{1}{u}$ substitutes into the denominator of the original integral and give you $\sqrt{1+u^2}$? Why is it not $\sqrt{1+((\frac{1}{u})^2)}$ ?
• Oct 31st 2007, 02:53 PM
galactus
Trig sub works well with this one. You have the correct integral in terms of trig functions.

I will use x instead of theta for less typing.

$\int\frac{sec{\theta}}{tan^{2}{\theta}}d{\theta}$

Rewrite as $\int{csc(x)cot(x)}dx$

Let $u=csc(x), \;\ du=-csc(x)cot(x)dx$

That gives the easiest of integrals:

$-\int{du}$

$-u$

$\boxed{-csc(x)}$
• Oct 31st 2007, 05:28 PM
Got5onIt
Thanks Galactus! I was hoping the fraction could be rewritten in some other trig form. I did not know it was cscXcotX. Where can I find a reference for trig subs? :)

Quote:

Originally Posted by galactus
Trig sub works well with this one. You have the correct integral in terms of trig functions.

I will use x instead of theta for less typing.

$\int\frac{sec{\theta}}{tan^{2}{\theta}}d{\theta}$

Rewrite as $\int{csc(x)cot(x)}dx$

Let $u=csc(x), \;\ du=-csc(x)cot(x)dx$

That gives the easiest of integrals:

$-\int{du}$

$-u$

$\boxed{-csc(x)}$