Solving Difficult Integral: Strategies & Tips

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Homework Statement


The whole integral is a double integral, but I can't even perform the first integration.

The integrand is cosx*sqrt(3+(cosx)^2) dxdy
The x bounds are from arctan(y) to pi/4
the y bounds are from 0 to 1

Homework Equations


Converting rectangular to polar may help.

The Attempt at a Solution



I've tried parts, but it was a disaster. I also started converting it to polar, but that was also a disaster.
 
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I switched dx and dy and the bounds.

The new y bounds are from 0 to tanx
The new x bounds are from 0 to pi/4

The resulting integrand after successfully integrating in terms of y is

-sinx*sqrt(3 + (cosx)^2) dx

It's a little nicer, but still ugly.

Putting the integrand into the mathematica online integrator, I get this:

(-3*ArcSinh[Cos[x]/Sqrt[3]])/2 - (Cos[x]*Sqrt[3 + Cos[x]^2])/2

How do I get here? I tried by starting with parts, but it doesn't get me far.
 
Solved with trig substitution and then parts.
 
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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