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I attached the papers...

Thanks in advance
 

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Speaking only for myself, if you want someone to take the time to look over what you did, at least post a photo that isn't rotated by 90°.

Even better would be if you posted the work directly in the input form.
 
Mark44 said:
Speaking only for myself, if you want someone to take the time to look over what you did, at least post a photo that isn't rotated by 90°.

Even better would be if you posted the work directly in the input form.

I actually tried to read it but my neck hurts while doing it, and I won't bother to download your photo and rotate it 90 degrees backwards to read it.
 
Millennial said:
I actually tried to read it but my neck hurts while doing it, and I won't bother to download your photo and rotate it 90 degrees backwards to read it.

I noticed that it was rotated after I started this thread. I actually did try to rotate it but didn't know how...anyways thanks for trying to read. I'll probably ask my professor about it.
 
Last edited:

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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