Integration involving a square root function.

NewtonianAlch
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Homework Statement


Integrate:

sqrt(1/4 + t^2 + t^4)

The Attempt at a Solution



I'm really not sure on how to go about integrating this, it's actually integrate from -1 to 1, the solutions manual has a method I'm not familiar with. I thought of factorising it first, although doing that hasn't made it any easier.
 
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I stared at it blankly for a little bit. But then I asked wolfram for the steps and said aha, why didn't I see that. If you pull the 1/4 out, you should recognize that 4t^4+4t^2+1 is the square of 2t^2+1. So you are integrating |2t^2+1|/2. But the absolute value goes away since 2t^2+1 is never negative.
 
I saw that in Wolfram too, although I didn't quite understand what was happening there.

So are you saying that pulling out the 1/4 from a square root doubles it when you pull it out? Or any other number for that matter.
 
Hmm, I think I see what you mean now, since the root of 1/4 is 0.5 you can take that out.

Interesting, doubt I'd have figured out all that myself without some help, thanks.
 
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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