Nasty Integral - Help with Trig Substitution

[mrworf]

Homework Statement

<br /> \int_{-p}^{p} \frac{2p}{(1+v^2)\sqrt{p^2 + v^2 +1 }} dv<br />

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:
<br /> \int_{-p}^{p} \frac{2p}{(1+v^2)\sqrt{\alpha^2+ v^2 }} dv<br />

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

we now have:
<br /> 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<br />

<br /> 2p \int_{-p}^{p} \frac{1}{(1+\alpha^2 \tan^2{\theta})\alpha \sec{\theta}} \alpha \sec{\theta}^2 d\theta<br />

<br /> 2p \int_{-p}^{p} \frac{ \sec{\theta}}{(1+\alpha^2 \tan^2{\theta})} d\theta<br />

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:<br /> \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 !