How to Integrate cos4 x Using the Double Angle Formula

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Homework Statement



Find the integral of cos4 xdx

Homework Equations



cos2x = 2cos2x-1
cos2x = (cos2x+1)/2

The Attempt at a Solution



I tried using cos4x = cos2x * cos2x and i simplified it to
cos4x = ((cos2x)2 + 2cos2x + 1)/4

I'm not sure if this is right. How do i go about getting the integral of this expression if it is correct. Thanks
 
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So far, so good. You just need to take it one step farther, by replacing (cos2x)2 with (cos4x + 1)/2. At that point you'll have an integral with four terms, and you can split this up into four integrals, each of which is pretty easy.
 
Hey i got the answer (sin4x)/32 + (sin2x)/4 + (3x)/8 + C which seems to be right. Thanks for your 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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