How to integrate $4\cos^2 (\theta) \sin^2 (\theta)$

Question

Should I change it into $4 \cos^2 (\theta) (1-\cos^2(\theta))$ or is there any possible way to use product rule?

Answer 1

Hint: Use that $$4\sin^2(x)\cos^2(x)= (\sin(2x))^2$$

Answer 2

TIP$_1$: $4\sin^2\theta \cos^2\theta = (2\sin\theta\cos\theta)^2 = \sin^2\,2\theta$.

TIP$_2$: $\cos\,2\theta = \cos^2 \theta - \sin^2\theta = 1 - 2\sin^2\theta$, for all $\theta \in \Bbb R$.

Answer 3

$$ \int { 4\cos ^{ 2 }{ x } \sin ^{ 2 }{ x } dx } =\int { { \sin ^{ 2 }{ \left( 2x \right) dx } } } \\ \int { \frac { 1-\cos { \left( 4x \right) } }{ 2 } } dx=\frac { x }{ 2 } -\frac { \sin { \left( 4x \right) } }{ 8 } +c $$