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.

The indefinite integral of your function is given by $$-{\frac {\ln  \left( -\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}\tan
 \left( x \right) +\sqrt {2}\sqrt {3}+2\, \left( \tan \left( x
 \right)  \right) ^{2} \right) \sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2
}}{120}}+{\frac {\ln  \left( -\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}
\tan \left( x \right) +\sqrt {2}\sqrt {3}+2\, \left( \tan \left( x
 \right)  \right) ^{2} \right) \sqrt {3}\sqrt {2\,\sqrt {2}\sqrt {3}-3
}}{60}}-1/30\,{\frac {2\,\sqrt {2}\sqrt {3}-3}{\sqrt {4\,\sqrt {2}
\sqrt {3}+6}}\arctan \left( {\frac {-\sqrt {2\,\sqrt {2}\sqrt {3}-3}
\sqrt {2}+4\,\tan \left( x \right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}}
 \right) }+1/30\,{\frac { \left( 2\,\sqrt {2}\sqrt {3}-3 \right) 
\sqrt {2}\sqrt {3}}{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {
\frac {-\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x
 \right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}} \right) }+1/3\,{\frac {
\sqrt {2}\sqrt {3}}{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {
\frac {-\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x
 \right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}} \right) }+{\frac {\ln 
 \left( \sqrt {2}\sqrt {3}+\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}
\tan \left( x \right) +2\, \left( \tan \left( x \right)  \right) ^{2}
 \right) \sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}}{120}}-{\frac {\ln 
 \left( \sqrt {2}\sqrt {3}+\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}
\tan \left( x \right) +2\, \left( \tan \left( x \right)  \right) ^{2}
 \right) \sqrt {3}\sqrt {2\,\sqrt {2}\sqrt {3}-3}}{60}}-1/30\,{\frac {
2\,\sqrt {2}\sqrt {3}-3}{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan
 \left( {\frac {\sqrt {2\,\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan
 \left( x \right) }{\sqrt {4\,\sqrt {2}\sqrt {3}+6}}} \right) }+1/30\,
{\frac { \left( 2\,\sqrt {2}\sqrt {3}-3 \right) \sqrt {2}\sqrt {3}}{
\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {\frac {\sqrt {2\,
\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x \right) }{\sqrt {4\,
\sqrt {2}\sqrt {3}+6}}} \right) }+1/3\,{\frac {\sqrt {2}\sqrt {3}}{
\sqrt {4\,\sqrt {2}\sqrt {3}+6}}\arctan \left( {\frac {\sqrt {2\,
\sqrt {2}\sqrt {3}-3}\sqrt {2}+4\,\tan \left( x \right) }{\sqrt {4\,
\sqrt {2}\sqrt {3}+6}}} \right)}
$$
solvable by the tan-half angle substitution.
