The answer is supposedly
$$\frac {1}{\sqrt{6}} \arctan\left(\sqrt{\frac {2}{3}}\tan x\right) + C$$
So I need to get it into form
$$\int \frac{\mathrm dx}{a^2+x^2} $$
but I am not sure what identities I need to use.
The answer is supposedly
$$\frac {1}{\sqrt{6}} \arctan\left(\sqrt{\frac {2}{3}}\tan x\right) + C$$
So I need to get it into form
$$\int \frac{\mathrm dx}{a^2+x^2} $$
but I am not sure what identities I need to use.
Hint: You can try dividing the numerator and denominator by $\cos^2 x$ and then enforcing the substitution $t=\tan x$ afterwards.
$$\int\frac {\mathrm dx}{2\sin^2 x+3\cos^2 x}=\int\mathrm dx\,\frac {\sec^2 x}{3+2\tan^2 x}=\int\frac {\mathrm dt}{3+2t^2}$$
I will let you complete the rest!