No definite integrals of trigonometry

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.

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.