Integrating cos^6(x) using cos^4θ - sin^4θ and cos^4θ + sin^4θ

[synkk]

Using $cos^4\theta - sin^4\theta = cos2\theta$ and $cos^4\theta + sin^4\theta = 1 - \frac{1}{2}sin^22\theta$

evaluate:

(i) $\displaystyle \int_0^{\frac{\pi}{2}} cos^4\theta \ d\theta$

adding the two identities given I get: $2cos^4\theta = cos2\theta + 1 - \frac{1}{4} + \frac{1}{4} cos4\theta$ $cos^4\theta = \frac{1}{8}\left (4cos2\theta + 3 + cos4\theta \right )$

integrating this I get the correct answer

(ii) $\displaystyle \int_0^{\frac{\pi}{2}} cos^6\theta \ d\theta$

from the (i) $cos^4\theta = \frac{1}{8}\left( 4cos2\theta + 3 + cos4\theta \right )$

so $cos^6\theta = \frac{1}{8}(4cos^2\theta cos(2\theta) +3cos^2\theta + cos^2\theta cos4\theta)$

using the double angle formula I simplify down to:
$cos^6\theta = \left(4\left( \dfrac{cos2\theta + 1}{2} \right)cos2\theta + 3\left(\dfrac{cos2\theta + 1}{2} \right) + cos^2\theta cos4\theta \right)$
$cos^6\theta = \left(cos4\theta + 1 + \frac{7}{2} cos2\theta + \frac{3}{2} + cos^2\theta cos4\theta \right )$
considering $cos^2\theta cos4\theta$

$= cos^2\theta(2(2cos^2\theta - 1)^2 - 1) = 8cos^6\theta - 8 cos^4\theta + cos^2\theta$

so:
$cos^6\theta = \frac{1}{8}\left(cos4\theta + \frac{5}{2} + \frac{7}{2}cos2\theta + 8cos^6\theta - 8 cos^4\theta + cos^2\theta \right)$

then the $cos^6\theta$ cancel out :S, any help?

(I've solved this problem using the reduction formula and de moivres theorem but I don't see where I'm going wrong here)

 

[mfb]

You can use integration by parts twice. You will get cos^6 on both sides there, too, but they don't cancel.
Alternatively, keep cos(4θ) as it is and play around with cos^2(θ).