I have the following integration to solve.

$$f(k) = \int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta,\quad0<k<1$$

assuming $\sin\theta = t$ which results $d\theta = \frac{dt}{\sqrt{1-t^2}}$ and when $\theta = 0, t=0$ and $\theta=\frac{\pi}{2},t=1$ so above equation can be rewritten as,

$$f(k) = \int_0^1{t^2\frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}}dt$$ I'm stuck in solving this further. Can somebody help me with some clues/solution to solve this further.

  • $\begingroup$ what is $k$? are there any conditions for $k$? $\endgroup$ – Jane Feb 7 '16 at 19:21
  • $\begingroup$ $f$ does not depend on $\theta$, so it should be $f(k)$ $\endgroup$ – Yuriy S Feb 7 '16 at 19:22
  • $\begingroup$ @Jane Thanks for your comment. condition for $k$ is $0<k<1$. $\endgroup$ – user3563929 Feb 7 '16 at 19:26
  • $\begingroup$ @YuriyS. Corrected. Thanks $\endgroup$ – user3563929 Feb 7 '16 at 19:28
  • $\begingroup$ You can see the final solution in terms of elliptic integrals here $\endgroup$ – Yuriy S Feb 7 '16 at 19:43

I will use $I(k)$ for the integral instead of $f(k)$.

$$ I(k)=\int^1_0 t^2 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}}dt $$

First, let's find some particular value, we will need it later.

$$I(1)=\frac{1}{3} $$

Now the definition for the elliptic integral of the second kind:

$$ E(k)=\int^1_0 \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}}dt $$

It's easy to show that:

$$ I(k)=E(k)-\int^1_0 \sqrt{1-t^2} \sqrt{1-k^2 t^2} dt $$

Taking $k$ derivative:

$$ \frac{dI}{dk}=\frac{dE}{dk}+k \int^1_0 t^2 \frac{\sqrt{1- t^2}}{\sqrt{1-k^2t^2}}dt $$

Now let's use integration by parts for $I(k)$:

$$ I(k)=-k^2 \int^1_0 t^2 \frac{\sqrt{1- t^2}}{\sqrt{1-k^2t^2}}dt+\int^1_0 \sqrt{1-t^2} \sqrt{1-k^2 t^2} dt $$

Finally we use all three equations to show:

$$ I(k)=-I(k)+E(k)-k\frac{dI}{dk}+k\frac{dE}{dk} $$

We get a linear ODE for $I(k)$:

$$ \frac{dI}{dk}=-\frac{2}{k} I(k)+\frac{dE}{dk}+\frac{E(k)}{k} $$

Using the usual method for such equations (reference here) we get the general solution:

$$ I(k)=\frac{C_1}{k^2}+\frac{1}{k^2} \int k^2 \left(\frac{dE}{dk}+\frac{E(k)}{k} \right) dk $$

To calculate the integral we use the known formula (can be seen here):

$$ \int k E(k) dk=\frac{1}{3} \left[(1+k^2)E(k)-(1-k^2)K(k) \right] $$

Integrating by parts:

$$ \int k^2 \left(\frac{dE}{dk}+\frac{E(k)}{k} \right) dk=k^2 E(k)-\int k E(k) dk $$

Finally, the general solution:

$$ I(k)=\frac{C_1}{k^2}+\frac{-(1-2k^2)E(k)+(1-k^2)K(k)}{3k^2} $$

Now we use the value $I(1)$ we calculated earlier and the known values $E(1)=1$ and $\lim_{k \rightarrow 1} (1-k^2)K(k)=0$ (see here) to obtain the final solution:


We can also check the result. From the original integral we can see that:


From the solution (and using series expansions for $E$ and $K$) we get:


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Possible hints to perfume some kind of calculations.

Leaving apart the extrema of the integra, for the moment.

$$\int\sin^2\theta \sqrt{1 - k^2 \sin^2\theta}\ \text{d}\theta$$

Using the substitution

$$k\sin\theta = \cos\phi ~~~~~~~ \sin\theta = \frac{\cos\phi}{k} ~~~ \to ~~~ \sin^2\theta = \frac{\cos^2\phi}{k^2}$$

$$\phi = \arccos(k\sin\theta)$$

$$\text{d}\phi = \frac{- k\cos\theta}{\sqrt{1 - k^2\sin^2\theta}}\ \text{d}\theta = -\frac{k\sqrt{1 - \sin^2\theta}}{\sqrt{1 - \cos^2\phi}}\ \text{d}\theta = -\frac{k\sqrt{1 - \frac{\cos^2\phi}{k^2}}}{\sin^2\phi}\ \ \text{d}\theta$$

Thence $$\text{d}\theta = -\frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\ \text{d}\phi$$ and the integral becomes

$$- \int \frac{\sin^2\phi}{\sqrt{k^2 - \cos^2\phi}}\left(\frac{\cos^2\phi}{k^2}\right)\sin^2\phi \text{d}\phi = -\frac{1}{k^2}\int \frac{\sin^4\phi\cos^2\phi}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi$$

Now we can use the trigonometric reduction formula for the numerator of the integrand:

$$\sin^4\phi\cos^2\phi = \frac{1}{32}(2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi))$$

to get

$$ -\frac{1}{32k^2}\int\ \frac{2 - \cos(2\phi) - 2\cos(4\phi) + \cos(6\phi)}{\sqrt{k^2 - \sin^2\phi}}\ \text{d}\phi $$

Which might be splitter into four parts, and then.. who knows!

The solution, however, lies into Jacobi Elliptic Functions of the First an Second Kind.

More on Jacobi Elliptic Integrals


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  • 1
    $\begingroup$ You made a mistake when calculating $d \phi$, there should be $-k \cos \theta$ in the numerator, which will complicate the integral $\endgroup$ – Yuriy S Feb 7 '16 at 20:25
  • $\begingroup$ @YuriyS Thant's it!!! Thank you so much, I'm going to delete/edit the answer! I don't think I will be helpful but who knows.. Upvote for you $\endgroup$ – Sigma Algebra Feb 7 '16 at 20:26
  • $\begingroup$ Note, that I made a mistake myself, I edited the comment. Check it again to be sure $\endgroup$ – Yuriy S Feb 7 '16 at 20:27
  • $\begingroup$ @YuriyS Haha funny! Don't worry, I found my error the same, thanks to your hint! $\endgroup$ – Sigma Algebra Feb 7 '16 at 20:28

Following on from @Yuriy-s, I have noticed that the integral:

$$\int k E(k) dk=\frac{1}{3} \left[(1+k^2)E(k)-(1-k^2)K(k) \right]$$

Is incorrect according to wolfram alpha. @Yuriy-s has used this to evaluate the expression, but note that wolfram alpha read this as:

$$\int k E(k^2) dk$$

Making a correction to this gives me this:

$$\int k E(k) dk=\frac{2}{45} \left[3k^2 + k-4)K(k)+(9k^2+k+4)E(k) \right] + C_2$$

where $C_2$ is a constant.

The following @Yuriy-s and subbing in various things:

$$I(k)=\frac{C_1}{k^2}+\frac{1}{k^2} \int k^2 \left(\frac{dE}{dk}+\frac{E(k)}{k} \right) dk$$

$$I(k)=\frac{C_1}{k^2}+E(k) - \frac{2}{45k^2} [(3k^2 +k -4) K(k) +(9k^2 +k +4)E(k)] + C_2$$

But there is a problem.

putting in the boundary conditions of: $$I(0)=\frac{\pi}{4}$$

$$E(0)=\frac{\pi}{2}$$ $$K(0)=\frac{\pi}{2}$$

Gives us: $$\frac{\pi}{4} = \frac{C_1}{0}+\frac{\pi}{2} - \frac{2}{0} [-4\frac{\pi}{2} +4\frac{\pi}{2}] +C_2 $$

$$C_2=\frac{-\pi}{4}$$ $$C_1=0$$

for a final result of: $$I(k)=E(k) - \frac{2}{45k^2} [(3k^2 +k -4) K(k) +(9k^2 +k +4)E(k)] - \frac{\pi}{4}$$

Unfortunately this is not the same as wolframs' final answer though:

$$I(k)=\frac{ \sqrt{1-k^2} [K(\frac{k^2}{k^2-1}) + (2k^2-1)E(\frac{k^2}{k^2-1}) ]}{3k^2}$$

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  • $\begingroup$ There are two typical conventions for elliptic integrals: Taking them to be functions of $k$, or as functions of $m=k^2$. Mathematica, and therefore Wolfram Alpha, uses the latter; the answer by Yuriy-s, by contrast, uses the former convention. That'd be my guess for why there's a seeming discrepancy. $\endgroup$ – Semiclassical Sep 1 '16 at 16:02
  • $\begingroup$ @Semiclassical-19 That seems plausible. But then why does wolfram give different results depending on whether you input the k varaible as any other symbol? wolframalpha.com/input/?i=integrate+a+EllipticE%5Ba%5D wolframalpha.com/input/?i=integrate+k+EllipticE%5Bk%5D $\endgroup$ – user209848 Sep 2 '16 at 11:47

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