How do I go about solving this equation for time?

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\ \frac{d^2\theta}{dt^2}=\frac{g\sin\theta-\frac{L}{2}\frac{d\theta}{dt}\cos2\theta}{\frac{I_{G}}{mL/2}+\frac{L}{2}\sin2\theta}

Solve for change in t between {\theta}=0 and {\theta}={\frac{\pi}{2}}

Any ideas?

Variables "g, L, I, m" are placeholders for the user to input values. Thanks so much!
 
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mcovalt said:
\ \frac{d^2\theta}{dt^2}=\frac{g\sin\theta-\frac{L}{2}\frac{d\theta}{dt}\cos2\theta}{\frac{I_{G}}{mL/2}+\frac{L}{2}\sin2\theta}

Solve for change in t between {\theta}=0 and {\theta}={\frac{\pi}{2}}

Any ideas?

Variables "g, L, I, m" are placeholders for the user to input values. Thanks so much!

Use tex and itex, not latex.
 
Oh man. That's weird. I swear it was fine when I posted it haha. Well, now that you can see it. Any ideas?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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