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

Let $$I = \int\frac{1}{\sin^6 x+\cos^6 x}dx = \int\frac{1}{(\sin^2 x+\cos^2 x)(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x)}dx$$
So $$\int\frac{1}{\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x}dx = \int\frac{1}{\sin^2 x\cos^2 x(\tan^2 x-1+\cot^2 x)}dx$$
So $$\int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x(\tan^2 x+\cot^2 x-1)}dx$$
So $$I = \int\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+1^2}dx$$
Now Put $(\tan x-\cot x) = t\;, $ Then $(\sec^2 x+\csc^2 x)dx = dt$
So $$I = \int\frac{1}{1+t^2}dt = \tan^{-1}(t)+\mathcal{C} = \tan^{-1}(\tan x-\cot x)+\mathcal{x}$$
