integral of sin^2 (x) * cos^2 (x) dx

[math]

Here's my attempt:

i know cos^2 x = cos2x + 1/2 (from double angle cos formula)

integral of (1 - cos^2 x) * cos^2 (x) dx = integral of cos^2 (x) - cos^4 (x)
integral of (cos2x / 2 + 1/2 ) - (cos(2x)/2 + 1/2)(cos(2x) / 2 + 1/2)
= integral of cos(2x) / 2 + 1/2 - cos^2 (2x) / 4 - cos(2x)/4 - cos(2x)/4 - 1/4
integral of -cos^2 (2x) / 4 + integral of 1/4

I'm stuck now.

Thanks.

 

[soroban]

Hello, math!

I would use: \(\displaystyle \:2\cdot\sin(x)\cdot\cos(x)\:=\:\sin(2x)\)

\(\displaystyle \L\int\)\(\displaystyle \sin^2x\cdot\cos^2x\,dx\)

\(\displaystyle \sin^2x\cdot\cos^2x \:=\:\frac{1}{4}\cdot\left(4\cdot\sin^2x\cdot\cos^2x)\:=\:\frac{1}{4}\cdot\left(2\cdot\sin x\cos x)^2 \:=\:\frac{1}{4}\cdot\sin^2(2x)\)

The integral becomes: \(\displaystyle \:\frac{1}{4}\L\int\)\(\displaystyle \sin^2(2x)\,dx \:=\:\frac{1}{4}\L\int \frac{1\,-\,\cos(2x)}{2}\,dx\)

And we have: \(\displaystyle \:\frac{1}{8}\L\int\)\(\displaystyle \left[1\,-\,\cos(2x)\right]\,dx\)

Got it?

 

[galactus]

\(\displaystyle \L\\\int{sin^{2}(x)cos^{2}(x)}dx\\=\frac{1}{4}\int[(1-cos(2x))(1+cos(2x))]dx\\=\frac{1}{4}\int(1-cos^{2}(2x))dx\\=\frac{1}{4}\int{sin^{2}(2x)}dx\\=\frac{1}{8}\int(1-cos(4x))dx\\=\frac{1}{8}x-\frac{1}{32}sin(4x)+C\)

 

[math]

yes, thank you galactus and soroban!