Definite integral of $1/(2\sin^4x + 3\cos^2x)$ I have $f = \frac 1 {(2\sin^4x + 3\cos^2x)}$ which area should be calculated from $0$ to $\frac{3\pi}2 $.
I noticed that $$\int_0^{\frac{3\pi}2} f \,dx= 3\int_0^{\frac{\pi}2} f \,dx$$
I tried to calculate this integral with the Weierstrass substitution:
$t = \tan{\frac x2}$
I got this integral:
$$6\int_0^1 \frac{(1+t^2)^3}{32t^4+3(1-t^4)^2} \,dt$$
My second try: I divided and multiplied on $\cos^4x$.
$$\int_0^{\frac{\pi}2} f \,dx = \int_0^{\frac{\pi}2} \frac1{(2\tan^4x + \frac3{\cos^2x})\cos^4x}  \,dx$$
$t = \tan x, dt = \frac{dx}{\cos^2x}, \cos^2x = \frac1{\tan^2 x +1}$
I got:
$$ \int_0^\infty \frac{t^2+1}{2t^4 + 3t^2 + 3} \,dt$$
Nothing of this helps me to calculate the area by getting primitive function.

Let me bother about $$I=\frac{1}{2}\int_{-\infty}^{\infty} \frac{1+1/t^2}{2t^2+\frac{3}{t^2}+3}dt.$$ Then
$$I=\frac{1}{2}\int_{-\infty}^{\infty} \frac{ A \left(\sqrt{2}-\frac{\sqrt{3}}{t^2}\right)+B \left(\sqrt{2}+\frac{\sqrt{3}}{t^2}\right)}{2t^2+\frac{3}{t^2}+3},~ A=\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{6}},~B=\frac{\sqrt{3}+\sqrt{2}}{2\sqrt{6}}.$$ Next, we re-write the denominator two ways: $$ I=\frac{A}{2} \int_{-\infty}^{\infty}  \frac{\left(\sqrt{2}-\frac{\sqrt{3}}{t^2}\right)}{\left(t\sqrt{2}+\frac{\sqrt{3}}{t} \right)^2-C^2} dt + \frac{B}{2} \int_{-\infty}^{\infty}  \frac{\left(\sqrt{2}+\frac{\sqrt{3}}{t^2}\right)}{\left(t\sqrt{2}-\frac{\sqrt{3}}{t} \right)^2+D^2} dt. $$ Here $C^2=2\sqrt{6}-3, D^2=2\sqrt{6}+3.$
Now let the twin transformations as  $u=(t\sqrt{2}+\sqrt{3}/t)$ and $v=(t\sqrt{2}-\sqrt{3}/t).$ Then $$I=A \int_{0}^{\infty} \frac{du}{u^2-C^2} + B  \int_{0}^{\infty} \frac{dv}{v^2+D^2}.$$ Thse are thw standard ones to be done next.
