No definite integrals of trigonometry

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $x+u=y\iff x=\cdots$

$\sin^2x=\dfrac{1-\cos2x}2=\dfrac{1-\cos2\left(y-u\right)}2$

$\cos2\left(y-u\right)=\cos2y\cos\left(2u\right)+\sin2y\sin\left(2u\right)$

$\cos\left(2u\right)=1-2\sin^2u=\cdots$

$\sin\left(2u\right)=2\sin u\cos u=\cdots$