Simplify the denominator in the following way:
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^2(2x)}{2}=\frac{1+\cos^2(2x)}{2}=\frac{2+\tan^2(2x)}{2\sec^2(2x)}$$
Hence, the integral you are dealing with is:
$$\int \frac{2\sec^2(2x)}{2+\tan^2(2x)}\,dx$$
I guess the next step is pretty obvious now. ;)
$$\sin^4x\cos^2x=\left(\frac{1-\cos2x}2\right)^2\frac{1+\cos2x}2$$
$$=\frac{(1+\cos2x)(1-2\cos2x+\cos^22x)}8$$
$$=\frac{1 -\cos2x-\cos^22x+\cos^32x}8$$
Again, $\cos^22x=\dfrac{1+\cos4x}2$
and $\cos3y=4\cos^3y-3\cos y\iff 4\cos^3y=\cos3y+3\cos y$ set $y=2x$
Alternatively use Euler Identities
$$2\cos x= e^{ix}+e^{-ix},2i\sin x=e^{ix}-e^{-ix}$$
$$(2i\sin x)^4(2\cos^2x)^2=(e^{ix}-e^{-ix})^4(e^{ix}+e^{-ix})^2$$
$$\implies64\sin^6x\cos^2x=(e^{ix}+e^{-ix})^2\cdot(e^{ix}+e^{-ix})^2(e^{ix}-e^{-ix})^2$$
$$=(e^{i2x}+e^{-i2x}+2)(e^{i2x}-e^{-i2x})^2$$
$$=(e^{i2x}+e^{-i2x}+2)(e^{i4x}+e^{-i4x}-2)$$
$$=e^{i6x}+e^{-i6x}-(e^{i2x}+e^{-i2x})+2(e^{i4x}+e^{-i4x})-4$$
$$=2\cos6x-(2\cos2x)+2(2\cos4x)-4$$
Best Answer
Hints:
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$