$\displaystyle \int\frac{tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$

$\displaystyle =2\int\frac{tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$

what should I do next?

Printable View

- Sep 6th 2009, 03:03 AMhonestliarI'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 AMHallsofIvy
- Sep 6th 2009, 03:37 AMhonestliar
- Sep 6th 2009, 04:07 AMchug1
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 AMANDS!
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 $\displaystyle \frac{1}{sec^{2}(\frac{x}{2})}$