No definite integrals of trigonometry I have big problems with the following integrals:
$$\int\frac{dx}{\sin^6 x+\cos^6x}$$
$$\int\frac{\sin^2x}{\sin x+2\cos x}dx$$
It isn't nice of me but I almost have no idea, yet I tried the trigonometric substitution $\;t=\tan\frac x2\;$ , but I obtained terrible things and can't do the rational function integral.
Perhaps there is exist some trigonometry equalities? I tried also
$$\frac1{\sin^6x+\cos^6x}=\frac{\sec^6x}{1+\tan^6x}=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\cdot\overbrace{\frac1{\sin^2x\cos^2x}}^{=\frac14\sin^22x}$$
and then doing parts with
$$u=\frac14\sin^22x\;\;:\;\;u'=\sin2x\cos2x=\frac12\sin4x\\{}\\v'=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\;\;:\;\;v=\arctan\tan^3x$$
But it is impossible to me doing the integral of $\;u'v\;$  .
Any help is greatly appreciated

$\bf{My\; Solution}::$ Given $$\displaystyle \int\frac{1}{\sin^6 x+\cos ^6 x}dx\;,$$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^6 x\;,$ 
We Get $$\displaystyle \int\frac{\sec^6 x}{1+\tan^ 6 x}dx\;,$$ Now let $$\tan x=t\;,$$ Then $$\sec^2 xdx = dt\;,$$ So we get
$$\displaystyle \int\frac{(1+t^2)^2}{1+t^6}dt = \int\frac{(1+t^2)^2}{(1+t^2)\cdot (t^4-t^2+1)}dt = \int\frac{1+t^2}{t^4-t^2+1}dt$$.
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $t^2\;,$ We get
$$\displaystyle \int\frac{\left(1+\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}-1}dt = \int\frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+1},$$ Now Let $$\displaystyle \left(t-\frac{1}{t}\right)=u\;,$$ Then $$\displaystyle \left(1+\frac{1}{t^2}\right)dt = du$$
So We get $$\displaystyle \int\frac{1}{u^2+1}du = \tan^{-1}\left(u\right)+\mathcal{C} = \tan^{-1}\left(\frac{t^2-1}{t}\right)+\mathcal{C}$$
So we get $$\displaystyle \int\frac{1}{\sin^6 x+\cos ^6x}dx = \tan^{-1}\left(\frac{\tan^2 x-1}{\tan x}\right)+\mathcal{C}$$
