Find the indefinite integral of $\int\frac{1}{(\frac{\cos^2x}{4}+\sin^2x)^2} \, dx$

Question

$$\int\frac{1}{(\frac{\cos^2x}{4}+\sin^2x)^2}\,dx$$ I do not know how to proceed, maybe is there any identity to make it easier to integrate?

Answer 1

The substitution $x\mapsto\arctan t$ leads to

$$ \mathfrak{I}=\int\frac{dt}{(1+t^2)\left(\frac{1}{4(1+t^2)}+\frac{t^2}{1+t^2}\right)^2}\,dt=\int\frac{(1+t^2)}{\left(\tfrac{1}{4}+t^2\right)^2}\,dt $$ which is an elementary integral. For any $a>0$ we have $$ \int\frac{dt}{a+t^2}=K_1+\frac{1}{\sqrt{a}}\,\arctan\frac{t}{\sqrt{a}} $$ and by applying $\frac{d}{da}$ to both sides we get: $$ \int\frac{dt}{\left(a+t^2\right)^2} = K_2+\frac{t}{2a(a+t^2)}+\frac{1}{2a\sqrt{a}}\,\arctan\frac{t}{\sqrt{a}} $$ from which: $$ \int\frac{1+t^2}{\left(\tfrac{1}{4}+t^2\right)^2}\,dt = \color{blue}{K_3+\frac{6t}{1+4t^2}+5\arctan(2t)}.$$