$$\int \frac{\sqrt{\cos 2x}}{1+\sin^2 x}\mathrm dx $$
I tried using $\cos 2x=1-2\sin ^2x$ and then putting $\sin x=t$ but it was of no use. I am running my mind on this problem since last two days but no success. Please help me with this problem.
Thanks!
EDIT -
I didn't know that this function doesn't have an elementary primitive. Thanks all for your help. You may leave this question and vote for closing.
This is an idea for solution.
1st method: The curve of $\frac{(Cos2x)^{0.5}}{1+Sin^2 x}$ intersect x and y axis at ($\pi$/4, 0)and (0, 1) respectively.To find limit of integral we first calculate the area under line passing point $(0, 1) and (\pi/4, 0)$ :
$A_1 =(\pi/4 . 1)/2=\pi/8$
$Cos 2x= 1-2 Sin^2 x$ and binomial expansion of $(Cos 2x)^{1/2}$ is:
$(1-2 Sin^2 x)^{1/2} = 1 - Sin^{2} x+(1/2) Sin^4 x -(1/2) Sin^6 x + . . .$
Dividing this by $(1+Sin^2 x)$ and intgrating the resulted polynomial we get:
$I=x -x/2 +(1/4)Sin 2x - ...$
For interval $[\pi/4, 0]$ we have:
$[x -x/2 +(1/4)Sin 2x - ...]^{\pi/4)}_{0} = \pi/8 + 1/4 + . . .$
2nd method: We can write: $$I = \int{\sqrt{Cos 2x}/{(1+ Sin^2 x)}}dx=\int[{\sqrt{Cos 2x}/2 Sin x Cos x}][{2 Sin x Cos x/(1+ Sin^2 x)]}dx $$
$\frac{\sqrt{Cos 2x}}{{(2 Sin x Cos x)}} = u$⇒ $du=\frac{(-1-{2 Cos^{3/2} 2x)}}{{(Sin^2 2x)}}dx$
$\frac{2 Sin x Cos x}{(1+ Sin^2 x)}dx = dv$ ⇒ $v =Ln {(1 + Sin^2 x)}$
⇒$I=\frac{\sqrt{Cos 2x}}{{Sin 2x}} . Ln {(1 + Sin^2 x)} + \int{[+1+\frac{2 Cos^{3/2} 2x}{Sin^2 2x}}]dx$
Or:
⇒$I=\frac{\sqrt{Cos 2x}}{{Sin 2x}} . Ln {(1 + Sin^2 x)} + x + \int{[\frac{2 Cos^{3/2} 2x}{Sin^2 2x}}]dx$
if $x=\pi/4$ ⇒$A=\frac{\sqrt{Cos 2x}}{Sin 2x} . Ln {(1 + Sin^2 x)} + x=\pi/4$
for $x=0$, A is infinity, but we find it for $\pi/8$ and we get about:
$A=\pi/8 + 0.56$
Therefore the value of integral is about $\pi/8 - 0.56 +...$ in interval $[\pi/4, \pi/8]$
Note that the fraction $\frac{(Cos2x)^{0.5}}{1+Sin^2 x}$is complex when $x=\pi/2$ and these calculations were in R.