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$$\int \frac{\sqrt{\cos 2x}}{1+\sin^2 x}\mathrm dx $$

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.

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    $\begingroup$ Who said it can be done in the first place? $\endgroup$ – Ivan Neretin Oct 16 '17 at 9:36
  • $\begingroup$ it leads to an elliptic integral $\endgroup$ – Dr. Sonnhard Graubner Oct 16 '17 at 9:38
  • $\begingroup$ Through the substitution $x=\arctan t$ it boils down to $$\int\frac{\sqrt{\frac{1-t^2}{1+t^2}}}{1+2t^2}\,dt \stackrel{t\mapsto\sqrt{s}}{=} \int\frac{\sqrt{\frac{1-s}{1+s}}}{2\sqrt{s}(1+2s)}\,ds\stackrel{\frac{1-s}{1+s}\mapsto u}{=}-\int\frac{\sqrt{u}}{\sqrt{1-u^2}(3-u)}\,du$$ but it does not get much better than that, it is an elliptic integral. Was the original problem about a definite integral? That may change things a bit. $\endgroup$ – Jack D'Aurizio Oct 16 '17 at 11:08
  • $\begingroup$ I'm voting to close this question as off-topic because I don't need any further answer as I haven't studied Elliptical Integral yet, and didn't know that this question was related to the same. $\endgroup$ – Jaideep Khare Oct 16 '17 at 11:35
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    $\begingroup$ Hint: The cures of $\frac{(Cos2x)^{0.5}}{1+Sin^2 x}$ and $\frac{Cos2x}{1+Sin^2x}$ intersect y and x axis at the same points ,for example($\pi$/4, 0), (0, 1).The difference is that $\frac{(Cos2x)^{0.5}}{1+Sin^2 x}$ is cave but $\frac{Cos2x}{1+Sin^2x}$ is concave. So the area under$\frac{(Cos2x)^{0.5}}{1+Sin^2 x}$ is the area between $\frac{Cos2x}{1+Sin^2x}$and x axis minus the area between these two curves.May be integral of$\frac{Cos2x}{1+Sin^2x}$ can give a rough approximation of integral of$\frac{(Cos2x)^{0.5}}{1+Sin^2 x}$. $\endgroup$ – sirous Oct 16 '17 at 12:41
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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.

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