Nasty Integral - Help with Trig Substitution

[mrworf]

Homework Statement

$\int_{-p}^{p} \frac{2p}{(1+v^2)\sqrt{p^2 + v^2 +1 }} dv$

Homework Equations

$1 + \tan{\theta}^2 = \sec{\theta}^2$

The Attempt at a Solution

I thought the best way to go about this was to rename some constants.
Let $\alpha^2 = 1 + p^2$ so that we are left with:
$\int_{-p}^{p} \frac{2p}{(1+v^2)\sqrt{\alpha^2+ v^2 }} dv$

i then make the variable substitution:
$v = \alpha \tan{\theta}$
$dv = \alpha \sec{\theta}^2 d\theta$

we now have:
$2p \int_{-p}^{p} \frac{1}{(1+\alpha^2 \tan^2{\theta})\sqrt{\alpha^2(1+\tan^2{\theta} )}} \alpha \sec{\theta}^2 d\theta$

$2p \int_{-p}^{p} \frac{1}{(1+\alpha^2 \tan^2{\theta})\alpha \sec{\theta}} \alpha \sec{\theta}^2 d\theta$

$2p \int_{-p}^{p} \frac{ \sec{\theta}}{(1+\alpha^2 \tan^2{\theta})} d\theta$

at this point i hit a roadblock - the (1 - \alpha \tan{\theta} ) term seems to be an inextricable pain . other substitutions leave me with a similar problem.
If anyone can point me in the direction of a better trig substitution, or perhaps an identity then your help would be greatly appreciated.
thanks for reading
-mrworf

 

[SammyS]

Homework Statement

$\int_{-p}^{p} \frac{2p}{(1+v^2)\sqrt{p^2 + v^2 +1 }} dv$

Homework Equations

$1 + \tan{\theta}^2 = \sec{\theta}^2$

The Attempt at a Solution

I thought the best way to go about this was to rename some constants.
Let $\alpha^2 = 1 + p^2$ so that we are left with:$\int_{-p}^{p} \frac{2p}{(1+v^2)\sqrt{\alpha^2+ v^2 }} dv$

i then make the variable substitution:
$v = \alpha \tan{\theta}$
$dv = \alpha \sec{\theta}^2 d\theta$

we now have:
$2p \int_{-p}^{p} \frac{1}{(1+\alpha^2 \tan^2{\theta})\sqrt{\alpha^2(1+\tan^2{\theta} )}} \alpha \sec{\theta}^2 d\theta$

$2p \int_{-p}^{p} \frac{1}{(1+\alpha^2 \tan^2{\theta})\alpha \sec{\theta}} \alpha \sec{\theta}^2 d\theta$

$2p \int_{-p}^{p} \frac{ \sec{\theta}}{(1+\alpha^2 \tan^2{\theta})} d\theta$

at this point i hit a roadblock - the (1 - \alpha \tan{\theta} ) term seems to be an inextricable pain . other substitutions leave me with a similar problem.
If anyone can point me in the direction of a better trig substitution, or perhaps an identity then your help would be greatly appreciated.
thanks for reading
-mrworf

Hello mrworf. Welcome to PF !

First of all, $\ \sec\theta^2\$ means $\ \sec(\theta^2)\,\$ whereas $\ \sec^2\theta\$ means $\ (\sec\theta)^2\ .\$ I'm sure you want the latter for this problem.

Also, you need to change the limits of integration to reflect the substitution you did.

 

[mrworf]

right on, with sammyS' suggestion, my equation becomes:

$2 p \int_{\tan^{-1}{\frac{p}{\alpha}}}^{\tan^{-1}{\frac{-p}{\alpha}}} \bigg[ \frac{\sec{\theta}}{(1+\alpha^2 \tan^2{\theta})} \bigg] d\theta$

 

[SammyS]

right on, with sammyS' suggestion, my equation becomes:

$\displaystyle 2 p \int_{\tan^{-1}{\frac{p}{\alpha}}}^{\tan^{-1}{\frac{-p}{\alpha}}} \bigg[ \frac{\sec{\theta}}{(1+\alpha^2 \tan^2{\theta})} \bigg] d\theta$

I admit, it's still quite nasty !