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

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.



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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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: $$I = \int{\sqrt{Cos 2x}/{(1+ Sin^2 x)}}dx=\int[{\sqrt{Cos 2x}/2 Sin x Cos x}][{2 Sin x Cos x/(1+ Sin^2 x)]}dx $$

$\frac{\sqrt{Cos 2x}}{{(2 Sin x Cos x)}} = u$⇒ $du=\frac{(-1-{2 Cos^{3/2} 2x)}}{{(Sin^2 2x)}}dx$

$\frac{2 Sin x Cos x}{(1+ Sin^2 x)}dx = dv$ ⇒ $v =Ln {(1 + Sin^2 x)}$

⇒$I=\frac{\sqrt{Cos 2x}}{{Sin 2x}} . Ln {(1 + Sin^2 x)} + \int{[+1+\frac{2 Cos^{3/2} 2x}{Sin^2 2x}}]dx$


⇒$I=\frac{\sqrt{Cos 2x}}{{Sin 2x}} . Ln {(1 + Sin^2 x)} + x + \int{[\frac{2 Cos^{3/2} 2x}{Sin^2 2x}}]dx$

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.