Trouble with Trigonometric Integral? Get Help Now!

Physter
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Haven't done integrals in such a long time and now I'm having some trouble with this question here. Any help would be appreciated. Thanks :smile:

http://img331.imageshack.us/img331/4333/screen192cj.jpg
 
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Using the double angle formula as you do here is of course legitimate, however, there is a simpler choice:
Substitute u=\sin(t) instead.
 
This looks better I hope?

http://img312.imageshack.us/img312/5109/screen202ul.jpg
 
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+C of course :). But yes, that's correct.
 
^^^Arggg I always forget the constant. Thanks o:)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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