Integration - Trig Substitutions (solve for Y)

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Homework Statement



Solve the problem for y as a function of x

a) (x^2+1)^2 \frac{dy}{dx}=\sqrt{x^2+1}

Homework Equations




The Attempt at a Solution


After some simplifying, I get here but get stuck:


Did I go the wrong route going for the U sub? Were my calculations wrong to begin with? Any help is greatly appreciated!
 
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You are on the right track. But the tex is hugely unclear. In the end the integral should have come out to be dtheta/sec(theta). And that's hugely easy.
 
Stop texing. Your integrand is 1/sec(t)^3 and the dx is sec(t)^2*dt.
 
Sorry guys. My texing was bad and it appears my question was too simple for these forums.
 
Hey, hey. Sorry. I wasn't criticizing you. I can't TeX for s**t either. I'm just saying the problem is easier than you think. You just didn't put the integrand together with the dx and get a simple solution.
 
Dick said:
Hey, hey. Sorry. I wasn't criticizing you. I can't TeX for s**t either. I'm just saying the problem is easier than you think. You just didn't put the integrand together with the dx and get a simple solution.

Well, I feel somewhat better. LOL!

Thanks for the help!
 
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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