- #1

mrworf

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## Homework Statement

[itex]

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

[/itex]

## Homework Equations

[itex] 1 + \tan{\theta}^2 = \sec{\theta}^2 [/itex]

## The Attempt at a Solution

I thought the best way to go about this was to rename some constants.

Let [itex]\alpha^2 = 1 + p^2 [/itex] so that we are left with:

[itex]

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

[/itex]

i then make the variable substitution:

[itex] v = \alpha \tan{\theta} [/itex]

[itex] dv = \alpha \sec{\theta}^2 d\theta [/itex]

we now have:

[itex]

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

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

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