As the title says, I need help with Integrate (sin(x))^2(cos(x))^2

I get thet I need to use the half- angle formula, but do I use it on both sin(x)^2 and cos(x)^2? And where do I go from there. Much appreciated!

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- May 1st 2009, 11:16 AMbirdzIntegrate (sin(x))^2(cos(x))^2
As the title says, I need help with Integrate (sin(x))^2(cos(x))^2

I get thet I need to use the half- angle formula, but do I use it on both sin(x)^2 and cos(x)^2? And where do I go from there. Much appreciated! - May 1st 2009, 11:24 AMReckoner
If you use the half-angle/power reduction formula on both factors from the start, you can then expand and use the formula again.

Alternatively, you could first use the Pythagorean identity, $\displaystyle \sin^2\theta+\cos^2\theta=1$ to get everything into terms of either sine or cosine. Then you could apply the half-angle formula twice. I don't think this would be any easier, but it is certainly valid. - May 1st 2009, 11:27 AMskeeter
$\displaystyle \left[\frac{1 - \cos(2x)}{2}\right] \cdot \left[\frac{1 + \cos(2x)}{2}\right] =$

$\displaystyle \frac{1 - \cos^2(2x)}{4} =$

$\displaystyle \frac{1}{4} \sin^2(2x) =$

$\displaystyle \frac{1}{4}\left[\frac{1 - \cos(4x)}{2}\right] =$

$\displaystyle \frac{1}{8}\left[1 - \cos(4x)\right]

$

integrate the last expression - May 1st 2009, 11:39 AMbirdz
Thank you for your very quick replies! I get it now :)

- May 2nd 2009, 01:07 AMMoo
It would have been better to directly note that $\displaystyle \sin(x)\cos(x)=\frac 12 \sin(2x)$, and then use the half angle formula (Nod)