Definite integral of $1/(2\sin^4x + 3\cos^2x)$

Question

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.

Answer 1

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:

$$3\frac{A}D\pi = 3\frac{\sqrt3+\sqrt2}{2\sqrt6\sqrt{2\sqrt6+3}}\pi$$

Answer 2

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.

Answer 3

To evaluate $\int_0^\infty\frac{t^2+1}{t^4+\frac32 t^2+\frac32}$ we'll first write $t^4+\frac32 t^2+\frac32=(t^2+at+b)(t^2-at+b)$ with $a:=\sqrt{\frac32-\sqrt{6}},\,b:=\sqrt{\frac32}$. Hence $$\int_0^\infty\frac{t^2+1}{(t^2+at+b)(t^2-at+b)}=\frac{1}{2ab}\int_0^\infty\sum_\pm\frac{a\pm (1-b)t}{t^2\pm at+b}dt\\=\frac{1}{2ab}\left[\sum_\pm\left(\pm\frac{1-b}{2}\ln(t^2\pm at+b)+\frac{a(1+b)}{\sqrt{4b-a^2}}\arctan\frac{2t\pm a}{\sqrt{4b-a^2}}\right)\right]_0^\infty\\=\frac{1+b}{2b\sqrt{4b-a^2}}\left[\sum_\pm\left(\frac{\pi}{2}\mp\arctan\frac{a}{\sqrt{4b-a^2}}\right)\right]\\=\frac{\pi(1+b)}{2b\sqrt{4b-a^2}}=\frac{\pi(2+\sqrt{6})\sqrt{2\sqrt{6}+1}}{6\sqrt{23}}.$$(You'll want to double-check my arithmetic at the end.)

Answer 4

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.