<s>[INST] 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.
 [/INST] Thank you, Dr Zafar Ahmed DSc, for a good idea, but your answer that I calculated is not correct. I tried to solve it myself with getting these coefficients (A, B, etc.). This is my solution that based on yours:
$$ \int_0^\infty \frac {1+\frac1{t^2}}{2t^2+\frac3{t^2}+3} \,dt$$
I. We can write $2t^2+\frac3{t^2}+3$ as 
$(\sqrt2t+\frac{\sqrt3}t)^2 - C^2$, where $C = 2\sqrt6 - 3$
or $(\sqrt2t+\frac{\sqrt3}t)^2 + D^2$, where $D = 2\sqrt6 + 3$
II. $1+\frac1{t^2}$ can be represented as $A*(\sqrt2+\frac{\sqrt3}{t^2}) + B(\sqrt2-\frac{\sqrt3}{t^2})$
We can find A and B by opening brackets and solving the system of two linear equations. 
$A = \frac{\sqrt3+\sqrt2}{2\sqrt6}, B = \frac{\sqrt3-\sqrt2}{2\sqrt6}$
III. Ok, now we can write our integral as:
$$ A\int_0^{\infty} \frac{\sqrt2 + \frac{\sqrt3}{t^2}}{(\sqrt2t-\frac{\sqrt3}{t})^2+D^2} \,dt + B\int_0^{\infty} \frac{\sqrt2 - \frac{\sqrt3}{t^2}}{(\sqrt2t+\frac{\sqrt3}{t})^2-C^2} \,dt$$
Look at the first integral in this sum:
$u = \sqrt2t-\frac{\sqrt3}{t}; du = \sqrt2+\frac{\sqrt3}{t^2}$
We have:
$$A\int_{-\infty}^{\infty} \frac{du}{u^2+D^2} = \frac{A}D\arctan{\frac{u}D}\Big|_{-\infty}^{\infty} = \frac{A}D\pi$$
Look at the second integral:
$u = \sqrt2t+\frac{\sqrt3}{t}; du = \sqrt2-\frac{\sqrt3}{t^2}$
We have
$$B\int_{\infty}^{\infty} \frac{dv}{v^2-C^2}=0$$
Then, the result is $$\frac{A}D\pi = \frac{\sqrt3+\sqrt2}{2\sqrt6\sqrt{2\sqrt6+3}}\pi$$
The answer to my based task is:
$$3\frac{A}D\pi = 3\frac{\sqrt3+\sqrt2}{2\sqrt6\sqrt{2\sqrt6+3}}\pi$$
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