integrate $\int \sin^{4}x\cos^{2}x$ 
$$\int \sin^{4}x\cos^{2}xdx$$

$$\int \sin^{4}x\cos^{2}xdx=\int (\sin x \cos x)^{2}\sin^2xdx=\int \left(\frac{\sin^{2}2x}{2}\right)\left(\frac{1}{2}-\frac{\cos2x}{2}\right)dx=\int \left(\frac{\sin^{2}2x}{4}-\frac{\sin^{2}2x\cos2x}{4}\right)dx=\frac{1}{4}\int ({\sin^{2}2x}-{\sin^{2}2x\cos2x})dx$$
I still have not mange to find $u$ substitution 
 A: $$\int\sin^4(x)\cos^2(x)\space\text{d}x=$$
$$\int\sin^4(x)\left(1-\sin^2(x)\right)\space\text{d}x=$$
$$\int\left(\sin^4(x)-\sin^6(x)\right)\space\text{d}x=$$
$$\int\sin^4(x)\space\text{d}x-\int\sin^6(x)\space\text{d}x=$$

You've to use twice the reduction formula:
$$\int\sin^m(x)\space\text{d}x=-\frac{\cos(x)\sin^{m-1}(x)}{m}+\frac{m-1}{m}\int\sin^{m-2}(x)\space\text{d}x$$

$$\frac{\sin^5(x)\cos(x)}{6}+\frac{1}{6}\int\sin^4(x)\space\text{d}x=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\int\sin^2(x)\space\text{d}x=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\int\left[\frac{1}{2}-\frac{\cos(2x)}{2}\right]\space\text{d}x=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{1}{2}\int1\space\text{d}x-\frac{1}{2}\int\cos(2x)\space\text{d}x\right]=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(2x)\space\text{d}x\right]=$$

Substitute $u=2x$ and $\text{d}u=2\space\text{d}x$:

$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(u)\space\text{d}u\right]=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(u)\space\text{d}u\right]=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{\sin(u)}{4}\right]+\text{C}=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{\sin(2x)}{4}\right]+\text{C}$$
A: $$\cos^2x = 1 - \sin^2x$$
Known this and the game is done. Then you have to compute
$$\int\sin^4(x) - \sin^6(x)\ \text{d}x$$
Which is quite easy. Can you proceed? 
Hint
Reduction formula
$$\int\sin^m(x)\ \text{d}x = -\frac{\cos(x) \sin^{n-1}(x)}{n} + \frac{n-1}{n}\int \sin^{n-2}(x)\text{d}x$$
A: Your own approach was easier than some of those suggested. Picking up from where you left off, you just use the double angle formula and reversing the chain rule, and we have $$\frac 14\int \frac 12(1-\cos 4x)dx-\frac 14\times \frac 16\sin^32x$$
Can you finish this now?
