1. ## Trig Integration help

So I've hit a wall, and as with other times I've posted, I'm really only looking for the &quot;kick-off&quot; step for this problem: $\displaystyle \int sin^2 x cos^2 x dx$ I have tried using the fact that the integrand is really a half-angle in disguise, but that leads to: $\displaystyle \int (1/2) sin^2 x dx$ which seems more complicated than the original problem. I've also tried to replace either the sin or cos term with 1+$\displaystyle cos^2 x$ etc., but that quickly makes the problem more complicated, or lands me back at square one. Thanks in advance for any and all help.

2. Originally Posted by flabbergastedman
So I've hit a wall, and as with other times I've posted, I'm really only looking for the &quot;kick-off&quot; step for this problem: $\displaystyle \int sin^2 x cos^2 x dx$ I have tried using the fact that the integrand is really a half-angle in disguise, but that leads to: $\displaystyle \int (1/2) sin^2 x dx$ which seems more complicated than the original problem. I've also tried to replace either the sin or cos term with 1+$\displaystyle cos^2 x$ etc., but that quickly makes the problem more complicated, or lands me back at square one. Thanks in advance for any and all help.
note the power reduction identity ... $\displaystyle \sin^2(u) = \frac{1-\cos(2u)}{2}$

$\displaystyle \sin^2{x}\cos^2{x} = \frac{1}{4}\sin^2(2x) = \frac{1 - \cos(4x)}{8}$

3. And that's the missing ingredient! Thank you very much