Integral: š›¼^(2x)cos(x)dx Solution

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Homework Statement


\inte^{2x}cos(x)dx

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The Attempt at a Solution


This isn't exactly a homework question, but it's close enough: it's from my old Calculus book (Larson), and I've been stuck on it for awhile. I just need a little hint or something. I've tried a u-substitution and haven't found anything that seems to help, and I've tried integration by parts, but nothing really simplifies--I just end up with an integral that is just as hard as before I try integration by parts. Maybe I'm using these techniques wrong or not seeing something obvious. No other 'harder' integration techniques should be needed (they come later in the book). Thanks!
 
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Try integrating by parts twice. You'll eventually end up with a term that looks like the original integral. You can collect terms, so to speak, and solve for the integral.
 
Ah! I would've never thought of that. 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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