Fairly basic integral failing to become solved. O:-)

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"Fairly" basic integral failing to become solved. O:-)

Hi,

I'm having trouble with an integral where I simply do not know where to start. I just need a little nudge in the right direction. I've tried integration by parts and by substitution, but I'm really just stumbling in the dark and should obviously choose something in a more well-informed manner. If I could just get a nudge in the right direction I think that I can solve it. :)

Homework Statement


\int (x^2+z^2)^{-3/2}dx
 
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when you see a square root in your integrand, you will likely want to attempt trig substitution if there is no other obvious route.

There are 3 different trig substitutions possible based on the form under the root:

x = z * sin@
x = z * tan@
x = z * sec@

you should know which one to use based on that form.

tell us which is the correct substitution, and you should be able to solve it.
 


I'll have a go at that, thank you very much.
 


Try a trig substitution, tan u = x/z. Keep in mind that z^2 is a constant in this integral.
 
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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