1. ## Integration

I need help with the following : int. cos²xsin²xdx
Thanks!

2. $\int \cos^2(x) \sin^2(x) \ dx = \int \bigg{(} \cos(x) \sin(x) \bigg{)}^2 \ dx$

now if we use the identity that $2 \cos(x) \sin(x) = \sin(2x)$

we have:

$\int \bigg{(} \frac{1}{2} \sin(2x) \bigg{)}^2 \ dx$ which should be simpler to integrate

3. Hello, nystudent2729!

Another substitution . . .

$\int \cos^2\!x \sin^2\!x\,dx$

$\int\cos^2\!x\sin^2\!x\,dx \;=\;\int\left(\frac{1+\cos2x}{2}\right)\left(\fra c{1-\cos2x}{2}\right)\,dx$

. . $=\;\frac{1}{4}\int(1-\cos^2\!2x)\,dx \;=\;\frac{1}{4}\int\sin^2\!2x\,dx$ .
. . . which is exactly what $^{|||||}$ got.

Then: . $\frac{1}{4}\int\left(\frac{1-\cos4x}{2}\right)\,dx \;=\;\frac{1}{8}\int(1 - \cos4x)\,dx$ . . . . etc.