General Solution for a Bernoulli Equation with Trigonometric Functions

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Homework Statement


Find the general solution of the following differential equation.
y'+4xy=10xy2cos(x2)

Homework Equations


The usual Bernoulli equation ones:
y'+p(x)y = g(x)ya
u(x)=[y(x)]1-a
u'+(1-a)pu = (1-a)g

The Attempt at a Solution


I got up until the general solution part.. I'll just type bits of it out because it'll take me ages (I'm a slow typer).

So in the equation, a=2, p(x)=4x, g(x)=10xcos(x2)

Change of variables:
u(x)=[y(x)]1-a = y-1

and u'+(1-a)pu = (1-a)g
=> u'-4xu = -10xcos(x2)

Now here's where I get confused..
General solution:
u= e\intp(x)dx[\intr(x)e\intp(x)dxdx+c
where p=-4x, r=-10xcos(x2)

u= e\int-2x^2[\int(-10xcos(x2)e\int-2x^2)dx+c

Argh.. that's messy, I hope it makes sense.

Anyway, I can't seem to figure out that integral... I've used parts and stuff but I don't seem to be getting anywhere.

Any help would be appreciated.
 
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Nevermind, I figured it out.
 
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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