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

HINT:
For the second one, as $\sin x+2\cos x=\sqrt5\sin\left(x+u\right)$ where $u=\arcsin\dfrac2{\sqrt5}\implies \sin u=\dfrac2{\sqrt5},\cos u=+\sqrt{1-\left(\dfrac2{\sqrt5}\right)^2}=\dfrac1{\sqrt5}$
let $x+u=y\iff x=\cdots$
$\sin^2x=\dfrac{1-\cos2x}2=\dfrac{1-\cos2\left(y-u\right)}2$
$\cos2\left(y-u\right)=\cos2y\cos\left(2u\right)+\sin2y\sin\left(2u\right)$
$\cos\left(2u\right)=1-2\sin^2u=\cdots$
$\sin\left(2u\right)=2\sin u\cos u=\cdots$
