integration - Integrate $int sin^4x cos^2x dx$

Integrate $$\int \sin^4x \cos^2x dx$$

Now, there's few solutions to this problem already on the internet. For example on yahoo:
https://answers.yahoo.com/question/index?qid=20090204203206AAbjUfM
and MIT
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-4-techniques-of-integration/part-a-trigonometric-powers-trigonometric-substitution-and-completing-the-square/session-69-integral-of-sin-n-x-cos-m-x-even-exponents/MIT18_01SCF10_ex69sol.pdf

I've tried a slightly different approach and simply wonder if my answer is correct as I don't know how to check it.

$$\begin{eqnarray} \int \sin^4x \cos^2x dx &=& \int \left( \frac{1-\cos^2x}{2} \right)^2 \cos^2x dx \\ &=& \frac{1}{4} \int (1-2\cos^2x+\cos^4x)\cos^2x dx \\ &=& \frac{1}{4} \int \cos^6x-2\cos^4x+\cos^2xdx \\ &=& \frac{1}{4} \left( \frac{1}{7}\cos^7x-\frac{2}{5}\cos^5x+\frac{1}{3}\cos^3x \right) + C \end{eqnarray}$$

Answer

$$\sin^4x\cos^2x=\left(\frac{1-\cos2x}2\right)^2\frac{1+\cos2x}2$$

$$=\frac{(1+\cos2x)(1-2\cos2x+\cos^22x)}8$$

$$=\frac{1 -\cos2x-\cos^22x+\cos^32x}8$$

Again, $\cos^22x=\dfrac{1+\cos4x}2$

and $\cos3y=4\cos^3y-3\cos y\iff 4\cos^3y=\cos3y+3\cos y$ set $y=2x$

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Alternatively use Euler Identities

$$2\cos x= e^{ix}+e^{-ix},2i\sin x=e^{ix}-e^{-ix}$$

$$(2i\sin x)^4(2\cos^2x)^2=(e^{ix}-e^{-ix})^4(e^{ix}+e^{-ix})^2$$

$$\implies64\sin^6x\cos^2x=(e^{ix}+e^{-ix})^2\cdot(e^{ix}+e^{-ix})^2(e^{ix}-e^{-ix})^2$$

$$=(e^{i2x}+e^{-i2x}+2)(e^{i2x}-e^{-i2x})^2$$

$$=(e^{i2x}+e^{-i2x}+2)(e^{i4x}+e^{-i4x}-2)$$

$$=e^{i6x}+e^{-i6x}-(e^{i2x}+e^{-i2x})+2(e^{i4x}+e^{-i4x})-4$$

$$=2\cos6x-(2\cos2x)+2(2\cos4x)-4$$