# I'm stuck, what should I do next?

• Sep 6th 2009, 03:03 AM
honestliar
I'm stuck, what should I do next?
$\displaystyle \int\frac{tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$
$\displaystyle =2\int\frac{tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$

what should I do next?
• Sep 6th 2009, 03:29 AM
HallsofIvy
Quote:

Originally Posted by honestliar
$\displaystyle \int\frac{tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$
$\displaystyle =2\int\frac{tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$

what should I do next?

First, what did you do? The two are obviously NOT equal. The second is twice the first.
• Sep 6th 2009, 03:37 AM
honestliar
Quote:

Originally Posted by HallsofIvy
First, what did you do? The two are obviously NOT equal. The second is twice the first.

I used u-substitution for the numerator.

should I change the denominator to $\displaystyle sec^{2}\frac{x}{2}$?
• Sep 6th 2009, 04:07 AM
chug1
Quote:

Originally Posted by honestliar
I used u-substitution for the numerator.

should I change the denominator to $\displaystyle sec^{2}\frac{x}{2}$?

Ah, well if you're using u-substitution it's especially important to specify the unit of integration at the end of your expression. The above confusion is evidence of what can go wrong when this isn't specified.

And off the top of my head, I think changing it to sec^2 would be a good idea, it might allow you to express the whole fraction as some form of sin^2 on cos^2 or vice versa.
• Sep 6th 2009, 08:10 AM
ANDS!
Quote:

Originally Posted by honestliar
I used u-substitution for the numerator.

should I change the denominator to $\displaystyle sec^{2}\frac{x}{2}$?

Yes. From there you can simplify your integral:

$\displaystyle \int\frac{tan(\frac{x}{2})}{sec^2(\frac{x}{2})}dx\ Rightarrow$

$\displaystyle \int\frac{sin(\frac{x}{2})cos^{2}(\frac{x}{2})}{co s(\frac{x}{2})}dx\Rightarrow$

$\displaystyle \int\sin(\frac{x}{2})cos(\frac{x}{2})dx$

From here, you can use u-du substitution. Do you see which one you should take?

From your earlier work I'm assuming you did:
$\displaystyle u=tan(\frac{x}{2})$

$\displaystyle du=\frac{1}{2}sec^{2}(\frac{x}{2})dx$

$\displaystyle 2du=sec^{2}(\frac{x}{2})dx$

Which is cool, except you don't have a $\displaystyle sec^{2}(\frac{x}{2})$ in your integral, you have a $\displaystyle \frac{1}{sec^{2}(\frac{x}{2})}$