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
 A: For the first integral, try simplifying the denominator as
$$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$$
$$=(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x$$
$$=1-\frac{3\sin^22x}{4}$$
This should start you off.
For the second integral, express $\sin x$ and $\cos x$ in terms of $\tan\left(\frac{x}{2}\right)$ and then substitute $u=\tan\left(\frac{x}{2}\right)$. Decompose into a lot of nasty looking partial fractions. For more help see here.
A: Hint: For the first integral, we can multiply by $\sec^{4}\left(x\right)
 $ numerator and denominator and get$$\int\frac{\sec^{2}\left(x\right)\left(\tan^{2}\left(x\right)+1\right)}{\tan^{4}\left(x\right)-\tan^{2}\left(x\right)+1}dx
 $$ and now take $u=\tan\left(x\right)
 $ $$\int\frac{u^{2}+1}{u^{4}-u^{2}+1}du=\int\frac{u^{2}+1}{\left(u^{2}-\frac{i\sqrt{3}}{2}-\frac{1}{2}\right)\left(u^{2}+\frac{i\sqrt{3}}{2}-\frac{1}{2}\right)}du
 $$ and now split using partial fractions. It gives a more tractable form. For the second integral take $u=\tan\left(\frac{x}{2}\right)
 $ to get $$-\int\frac{4u^{2}}{u^{6}-u^{5}+u^{4}-2u^{3}-u^{2}-u-1}du=-\int\frac{4u^{2}}{\left(u^{2}+1\right)^{2}\left(u^{2}-u-1\right)}du
 $$ and again you can use partial fractions. 
A: 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 $x+u=y\iff x=\cdots$
$\sin^2x=\dfrac{1-\cos2x}2=\dfrac{1-\cos2\left(y-u\right)}2$
$\cos2\left(y-u\right)=\cos2y\cos\left(2u\right)+\sin2y\sin\left(2u\right)$
$\cos\left(2u\right)=1-2\sin^2u=\cdots$
$\sin\left(2u\right)=2\sin u\cos u=\cdots$
A: $\bf{My\; Solution}::$ 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 $$\displaystyle \int\frac{1}{\sin^6 x+\cos ^6x}dx = \tan^{-1}\left(\frac{\tan^2 x-1}{\tan x}\right)+\mathcal{C}$$
A: 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 $$I = \int\frac{1}{1+t^2}dt = \tan^{-1}(t)+\mathcal{C} = \tan^{-1}(\tan x-\cot x)+\mathcal{x}$$
