Integral $\int\frac{\sin2x+2tanx}{\cos^{6}x+6\cos^{2}x+4}dx$

Question

Q) $\int\frac{\sin2x+2tanx}{\cos^{6}x+6\cos^{2}x+4}dx$

Tried to split it and integrate individual but , but getting an irreducible cubic equation in denominator . Any magic substitution ?

Answer 1

The substitution $u=\tan(x/2)$ will give a rational function, which in principle could be integrated. Here is another solution, that we are lucky to do. One could also in this case do $u=\tan x$ directly...

Using that $$ \sin 2x=2\sin x\cos x=2\tan x\cos^2 x=\frac{2\tan x}{1+\tan^2x} $$ and $$ \cos^6x+6\cos^2x+4=\frac{1+6(1+\tan^2x)+4(1+\tan^2x)^3}{(1+\tan^2x)^3}, $$ we get, after some expanding, $$ \frac{1}{12}\int \frac{48\tan x+72\tan^3x+24\tan^5x}{11+24\tan^2x+18\tan^4x+4\tan^6x}(1+\tan^2x)\,dx. $$ Here, we are really lucky(?). Since $$ D(11+24u^2+18u^4+4u^6)=48u+72u^3+24u^5, $$ and since $D\tan x=1+\tan^2x$, we find that $$ \begin{gathered} \frac{1}{12}\int \frac{48\tan x+72\tan^3x+24\tan^5x}{11+24\tan^2x+18\tan^4x+4\tan^6x}(1+\tan^2x)\,dx\\ =\frac{1}{12}\log(11+24\tan^2x+18\tan^4x+4\tan^6x)+C. \end{gathered} $$

Answer 2

Saying $\cos^2 x=u$ is best for me.

Moreover, uper part is $\sin 2x(1+\frac{1}{\cos^2 x})$.