Solving $\int_0^{\pi/2} \sin^2\theta \sqrt{1-k^2\sin^2 \theta}d\theta $

Question

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.

Answer 1

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:

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

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

$$I(0)=\frac{\pi}{4}$$

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

$$I(0)=\frac{\pi}{6k^2}(1-k^2+k^2/4-1+2k^2+k^2/4)=\frac{\pi}{4}$$

Answer 2

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

https://en.wikipedia.org/wiki/Elliptic_integral

Answer 3

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:

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

Answer 4

No I find a problem with the last formula: with MAPLE the real part is false but the imaginary part is right. A true formula is according to WOLFRAM is [(1-k^2)K(k^2)+(2k^2-1)E(k^2)]/3k^2.