I've been trying to solve this problem – $\int\cos^4(x)\sin^4(x)\ dx$, but I don't seem to be succeeding. This is what I've done: $$\int\left(\frac{1+\cos(2x)}{2}\right)^2\left(\frac{1-\cos(2x)}{2}\right)^2\ dx\\=\frac1{16}\int\left(1+2\cos(2x)+\cos^2(2x)\right)\left(1-2\cos(2x)+\cos^2(2x)\right)\ dx\\=\frac1{16}\int1+2\cos^2(2x)+\cos^4(2x)\ dx\\=\frac1{16}\int1+1+\cos(4x)+\frac{
(1+\cos(4x))^2}{4}\ dx\\=\frac1{16}\int2+\cos(4x)+\frac{1+2\cos(4x)+\cos^2(4x)}{4}\ dx\\=\frac1{16}\int2+\cos(4x)+\frac{1}{4}+\frac{\cos(2x)}{2}+\frac{1+\cos(8x)}{8}\ dx\\=\frac1{16}\left(2x+\frac{\sin(4x)}{4}+\frac{x}4+\frac{\sin(2x)}{4}+\frac{x}8+\frac{\sin(8x)}{64}\right)\\=\frac{x}{8}+\frac{\sin(4x)}{64}+\frac{x}{64}+\frac{\sin(2x)}{64}+\frac{x}{128}+\frac{\sin(8x)}{1024}\\=\frac{19x}{128}+\frac{\sin(2x)}{64}+\frac{\sin(4x)}{64}+\frac{\sin(8x)}{1024}$$
I checked this up on integral-calculator.com. They seem to have a similiar answer – but not quite the same – their denominators are different and they don't have a $\sin(2x)$ term. I keep retrying this problem, but I seem to be missing something. Where am I going wrong?
Solving: $\int\cos^4(x)\sin^4(x)\ dx$
calculusintegrationtrigonometry
Best Answer
Generally $$ (1+2a+b)(1-2a+b)\neq 1+2a^2+b^2, $$ as it is in your calculations (lines 2 and 3) with $a=\cos(2x)$ and $b=\cos^2(2x)$.
Probably a quicker way is to use $\sin2x=2\sin x\cos x$ and reduce to solving integral of the type $$ \int\sin^4x\,dx. $$