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The integral was:

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^2x}{(a^2+b^2\sin^2x)^{3/2}}\;dx= \frac{\pi}{2ab^2} (1-\frac{a}{\sqrt{a^2+b^2}}).$$

  • I encountered this integral while trying to show amperes law working in an EM (electromagntism) problem .

My progess: I tried substituting $a^2 + b^2 \sin^2x = t^2 $ but that doesnt help much. Next I tried writing $\cos^2x$ in terms of $\sin^2x$ and then separately do the integration for two parts in the expression but that I wasnt able to show. Any help greatly appreciated.

The setup was this (figure added ) enter image description here : a semi infinite wire straight current $I$ being given to a plane sheet ( current flows radially there ) .

  • Find magnetic field at a general point .

(By some symmetry reasons one compute it to be $B = \frac{u_°I}{2\pi r}$ $\hat{k}$ at a point $(r,h,0)$ by Amperes Law, the plane being $y=0$ and semi infinite line being the y axis , current direction in that in $-y$ direction .

Now by integration approach ( figure added ) enter image description here i first find the magentic field due to a current dI = $\frac{I}{2\pi} d\theta$ between $\theta$ and $\theta +d\theta$ angle in the plane at that point, it comes out to be

  • $\frac{u_°I}{4\pi(h^2 + r^2sin^2\theta)} (1+ \frac{rcos\theta}{\sqrt(h^2 + r^2 sin^2\theta)})$ ($hsin\theta \hat{i} - rsin\theta \hat{j} + hcos\theta \hat{k})$
  • Now integrating in limits $0$ to $2\pi$ the $\hat{i}$ and $\hat{j}$ components goes to zero and we are left with.
  • $\int_{0}^{\pi}\frac{u_°I}{4\pi(h^2 + r^2sin^2\theta)} (1+ \frac{rcos\theta}{\sqrt(h^2 + r^2 sin^2\theta)}) (2 hcos\theta \hat{k})$

This with the straight wire current magentic field which is standard integral ( means easily evalualted) should result in what we get from amperes law as net magnetic field at that point . On comparing we arrived at the result i mentioned at first . Now if there is any mistake in my approach that would also be fine as acceptable answer . If the calculation and approach is fine then only share the method for evaluating that integral .

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    $\begingroup$ I suppose that the exponent is $1$ and not $3/2$. Otherwise, you would face elliptic integrals. So, if the rhs is correct, the problem is simple $\endgroup$ Commented Oct 10, 2022 at 9:55
  • $\begingroup$ I arrived at that by finding the magnetic field at a point on the setup of a semi infinite straight wire current being given into a plane (line is perpendicular to plane), so current flows radially in that plane . Using integration approach and other by amperes law . @ClaudeLeibovici , the answer is dimensionally correct though i think $\endgroup$ Commented Oct 10, 2022 at 10:00
  • $\begingroup$ If this is your integral, the rhs is wrong $\endgroup$ Commented Oct 10, 2022 at 10:02
  • $\begingroup$ Should i share the approach i used in that EM problem to arrive at this ? In my question , that would be better i think @ClaudeLeibovici $\endgroup$ Commented Oct 10, 2022 at 10:04
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    $\begingroup$ Your formula is false $\endgroup$ Commented Oct 10, 2022 at 12:32

2 Answers 2

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As I commented earlier, the rhs corresponds to an exponent equal to $1$ and not $\frac 32$.

For the lhs, you face elliptic integrals. Concerning the antiderivative, if $0 \leq x \leq \frac \pi 2$, $$\int \frac{\cos^2(x)}{\left(a^2+b^2\sin^2(x)\right)^{\frac 32}}\,dx=\frac{E\left(x\left|-\frac{b^2}{a^2}\right.\right)-F\left(x\left|-\frac {b^2}{a^2}\right.\right)}{a b^2} +\frac{\sin (2 x)}{\sqrt{2} a^2 \sqrt{2 a^2-b^2 \cos (2 x)+b^2}}$$ where appear the incomplete elliptic integrals of the second and first kinds.

For the integral over the given bounds, the last term disappears and the results write

$$\int_0^{\frac \pi 2} \frac{\cos^2(x)}{\left(a^2+b^2\sin^2(x)\right)^{\frac 32}}\,dx=\frac{E\left(-\frac{b^2}{a^2}\right)-K\left(-\frac{b^2}{a^2}\right)}{a b^2}$$ where appears the complete elliptic integrals of the second and first kinds.

This answer is also dimensionally correct.

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  • $\begingroup$ Claude thanks for this answer but i was asking for the wrongess in the physics or maths which i did to arrived at that result of the integration . I want to know that. Not the integral value itself $\endgroup$ Commented Oct 12, 2022 at 9:33
  • $\begingroup$ I shared all the details of my calculation already above in post $\endgroup$ Commented Oct 12, 2022 at 9:33
  • $\begingroup$ you may just tell where in my method i should do something so that only 1 is in the exponent and not 3/2 ? Thats the mistake i am asking of $\endgroup$ Commented Oct 12, 2022 at 9:56
  • $\begingroup$ @ProblemDestroyer you should ask, how the exponent came 3/2, on physics.stackexchange.com $\endgroup$ Commented Oct 13, 2022 at 12:13
  • $\begingroup$ Alright sure i will $\endgroup$ Commented Oct 13, 2022 at 13:37
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$$I=\int\limits_0^{\large\frac\pi2}\dfrac{\cos^2x\,\text dx}{(a^2+b^2\sin^2x)^{\large\frac32}} =\dfrac1{b^2}\int\limits_0^{\large\frac\pi2}\dfrac{a^2+b^2-(a^2+b^2\sin^2x)}{(a^2+b^2\sin^2x)^{\large\frac32}}\,\text dx$$ $$=\dfrac{a^2+b^2}{b^2}\int\limits_0^{\large\frac\pi2}\dfrac{\text dx}{(a^2+b^2\sin^2x)^{\large\frac32}} -\dfrac1{b^2}\int\limits_0^{\large\frac\pi2}\dfrac{\text dx}{\sqrt{a^2+b^2\sin^2x}}.$$ Since $$\int\limits_0^{\large\frac\pi2}\dfrac{\text dx}{(a^2+b^2\sin^2x)^{\large\frac32}} = \dfrac1{|a|(a^2+b^2)}E\left(-\dfrac{b^2}{a^2}\right),$$ $$\int\limits_0^{\large\frac\pi2}\dfrac{\text dx}{\sqrt{a^2+b^2\sin^2x}} = \dfrac1{|a|}K\left(-\dfrac{b^2}{a^2}\right),$$ then $$I=\dfrac{E\left(-\dfrac{b^2}{a^2}\right)-K\left(-\dfrac{b^2}{a^2}\right)}{|a|b^2}.\tag1$$

Taking in account the series representations $$E(x)=\dfrac\pi2\left(1+\left(\dfrac12\right)^2\,x+\left(\dfrac12\cdot\dfrac34\right)^2\,x^2+\left(\dfrac12\cdot\dfrac34\cdot\dfrac56\right)^2\,x^3+\dots\right),\qquad |x|\le 1,$$ $$K(x)=\dfrac\pi2\left(1-\left(\dfrac12\right)^2\,\dfrac x1 -\left(\dfrac12\cdot\dfrac34\right)^2\,\dfrac{x^2}3-\left(\dfrac12\cdot\dfrac34\cdot\dfrac56\right)^2\,\dfrac{x^3}5-\dots\right),\qquad |x|\le 1,$$ easily to see that the obtained solution $(1)$ does not correspond with the expression $$\dfrac\pi{2ab^2}\left(1-\dfrac a{\sqrt{a^2+b^2}}\right),$$ proposed in OP.

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