# Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? [closed]

Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$

## closed as off-topic by Semiclassical, choco_addicted, heropup, Ben Sheller, M. VinayApr 21 '16 at 16:24

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• Divide top and bottom of the integrand by $a^2\cos^2\theta$ and note that the derivative of $\tan\theta$ is $\sec^2\theta$. If that is not enough, what is the derivative of $\tan^{-1}x$? – almagest Apr 21 '16 at 12:51
• @Derp Please show how you attempted to solve the problem and what you are stuck on. This way people won't downvote your question. You've been on this site for quite a while so you should know this by now. – Arbuja Apr 21 '16 at 14:07
• Possible duplicate of Find $\int_0^{2\pi} \frac{1}{a^2\cos^2 t+b^2 \sin^2 t} dt \;; a,b&gt;0$. – Ben Sheller Apr 21 '16 at 16:13
• For this question to be off topic in this forum seems rather like saying $2$-liter soda bottles are off topic for FPSRussia. Sure, you've seen it dispatched many times before, but there have been some creative approaches. I am up to a count of something like $13$ methods at this point. In this forum is it reasonable to have a master question that collects all the answers to an FAQ like this so it can be referenced in further queries? – user5713492 Apr 21 '16 at 18:42

Let $z=e^{i\theta}$, then \begin{align} \int_0^{2\pi}\frac{d\theta}{a^2\cos^2\theta+b^2\sin^2\theta}&=\int_C \frac{1}{a^2\left(\frac{z+\frac{1}{z}}{2}\right)^2+b^2\left(\frac{z-\frac{1}{z}}{2i}\right)^2}\frac{dz}{iz}\\ &=\int_C \frac{-4iz}{(a^2-b^2)z^4+2(a^2+b^2)z^2 +(a^2-b^2)}dz, \end{align} where $C:|z|=1$. If $a=b$, then the integral becomes $$\int_C \frac{-i}{a^2 z}dz = 2\pi i \operatorname{Res}\left(-\frac{i}{a^2 z};0\right)=\frac{2\pi}{a^2}$$ If $a\ne b$, Solve $(a^2-b^2)z^4+2(a^2+b^2)z^2+(a^2-b^2)=0$, then $$z^2=-\frac{a+b}{a-b}\text{ or }z^2=-\frac{a-b}{a+b}$$ by quadratic formula. Since $\left|\frac{a+b}{a-b}\right|>1$ and $\left|\frac{a-b}{a+b}\right|<1$, there exist two simple poles inside $C$. Assuming $a>b>0$, \begin{align} \int_C \frac{-4iz}{(a^2-b^2)z^4+2(a^2+b^2)z^2 +(a^2-b^2)}dz &= 2\pi i\left(\operatorname{Res}\left(f;\sqrt{\frac{a-b}{a+b}}i\right)+\operatorname{Res}\left(f;-\sqrt{\frac{a-b}{a+b}}i\right)\right) \end{align} Compute residues: \begin{align} \operatorname{Res}\left(f;\sqrt{\frac{a-b}{a+b}}i\right) &= \frac{4\sqrt{\frac{a-b}{a+b}}}{a^2-b^2}\frac{1}{2\sqrt{\frac{a-b}{a+b}}i\left(-\frac{a-b}{a+b} + \frac{a+b}{a-b}\right)}\\ &=\frac{2}{4iab}=\frac{1}{2abi} \end{align}

\begin{align} \operatorname{Res}\left(f;-\sqrt{\frac{a-b}{a+b}}i\right) &=\frac{-4\sqrt{\frac{a-b}{a+b}}}{a^2-b^2}\frac{1}{-2\sqrt{\frac{a-b}{a+b}}i\left(-\frac{a-b}{a+b} + \frac{a+b}{a-b}\right)}\\ &=\frac{1}{2abi} \end{align}

$$\therefore \int_C \frac{-4iz}{(a^2-b^2)z^4+2(a^2+b^2)z^2 +(a^2-b^2)}dz = 2\pi i \left(\frac{1}{2abi}+\frac{1}{2abi}\right)=\frac{2\pi}{ab}$$ We can get same conclusion when $b>a>0$.

• – 1 0 Apr 21 '16 at 14:40
• @MithleshUpadhyay Maybe it is just an error of the program. Given problem is contained in Fundamentals of Comlex Analysis, written by E. B. Saff, A. D. Snider, 3rd ed, page 317. – choco_addicted Apr 21 '16 at 14:56

The given integral equals to $$\\{{4}\over{b^2}}\int_0^{{\pi/2}}{{\sec^2\theta\\d\theta}\over{(a^2/b^2)+\tan^2\theta}}$$

put $\tan\theta = t$ which implies $$\\{{4}\over{b^2}}\int_0^{{\pi/2}}{{\\dt}\over{(a^2/b^2)+t^2}}$$ which after integaration equals to

$$\\{{4}\over{b^2}}*{{b}\over{a}}*{\pi\over2} = {{2\pi}\over{ab}}$$ I hope this was helpful.

$\cos^2\theta,\sin^2\theta$ have period $\pi$, so the integral is $2I$ where $I$ is the integral from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. wlog we may take $a,b$ to be positive.

We have $I=\frac{1}{a^2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sec^2\theta}{1+(\frac{b}{a})^2\tan^2\theta}\ d\theta$. Putting $x=\tan\theta$ this becomes $\frac{1}{a^2}\int_{-\infty}^{\infty}\frac{dx}{1+(\frac{b}{a})^2x^2}$. Putting $y=\frac{b}{a}x$ we get $\frac{1}{ab}\int_{-\infty}^{\infty}\frac{dy}{1+y^2}=\frac{\pi}{ab}$. [The last integral is just $\tan^{-1}y$.]

Hence the original integral is $\frac{2\pi}{ab}$ as required.

We use the fact that the 1-form

$$\eta = \frac{x dy - y dx}{x^2 + y^2}$$

has integral of $2 \pi$ over $\gamma_r(t) = (r \cos t, r \sin t)$.

Furthermore, if $\Gamma(0) = \Gamma(2\pi)$ and if the intervals $[\gamma(t), \Gamma(t)]$ do not contain $\mathbf{0}$ for any $t \in [0, 2 \pi]$, then the integral over $\Gamma$ is also zero.

Now take $\Gamma(t) = (a \cos t, b \sin t)$. We have

\begin{align} 2\pi = \int_{\Gamma}\eta &= \int_{0}^{2\pi} \frac{a \cos t}{a^2 \cos^2 t + b^2 \sin^2 t} b \cos t + \frac{-b \sin t}{a^2 \cos^2 t + b^2 \sin^2 t} (-a \sin t) \\ &=\int_0^{2\pi} \frac{ab}{a^2 \cos^2 t + b^2 \sin^2 t}. \end{align}

Proof of the statements stated above:

\begin{align} \int_{\gamma} \eta &= \int_0^{2\pi} \sum_{i=1}^2 a_i(\gamma(t)) \frac{\partial\gamma_i}{\partial t} \, dt\\ &= \int_0^{2\pi} -\frac{\sin t}{r} (-r \sin t) + \frac{\cos t}{r} r \cos t \, dt \\ &= \int_0^{2\pi} \sin^2 t + \cos^2 t \, dt = 2\pi.\end{align}

with $a_1 = \dfrac{-y}{x^2 + y^2}, a_2 = \dfrac{x}{x^2 + y^2}$.

Now, \begin{align*}d \eta &= (da_1) \wedge dx_1 + (da_2) \wedge dx_2\\ &= D_2 a_1 \, dx_2 \wedge dx_1 + D_1 a_2 \, dx_1 \wedge dx_2\\ &= ((D_1 a_2)(x, y) - (D_2a_1)(x, y)) \, \wedge dx_1 \wedge dx_2\\ &= \frac{y^2 - x^2}{(x^2 + y^2)^2} - \frac{y^2 - x^2}{(x^2 + y^2)^2} \, dx_1 \wedge dx_2 = 0. \end{align*}

Next, let $\Gamma$ be as described. Take $$\Phi(t, u) = (1-u)\Gamma(t) + u\gamma(t).$$

We get $\partial \Phi = \Gamma -\gamma$

Hence $$0 = \int_{\Phi} d\eta = \int_{d \Phi} \eta = \int_{\Gamma - \gamma} \eta$$

by Stokes' theorem

and so $$\int_\Gamma \eta = \int_\gamma \eta.$$

• This is really nuking a mosquito here, but I like the diversity of approaches – ASKASK Apr 21 '16 at 15:06
• This was a hard first read, but my understanding is that you said if $$\vec F(x,y,z)=\langle\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\rangle$$ then it's easy to compute $$\int_{x^2+y^2=a^2,z=0}\vec F\cdot d\vec r=2\pi$$ so you can use Stokes' theorem to get $$\int_{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,z=0}\vec F\cdot d\vec r$$ because in the area between the circle and the ellipse, $$\vec\nabla\times\vec F=\vec0$$ Thus the eighth method. From which +1 follows :) – user5713492 Apr 21 '16 at 15:20

The ninth method is an easy corollary to @Soke's answer. Consider the vector field $$\vec F(x,y)=\langle\frac x{x^2+y^2},\frac y{x^2+y^2}\rangle$$ and the path $\Gamma$ $$\vec r(\theta)=\langle a\cos\theta,b\sin\theta\rangle,\,0\le\theta\le2\pi$$ Along the path, $$d\vec r=\langle-a\sin\theta,b\cos\theta\rangle d\theta$$ So $$\hat n\,ds=\langle b\cos\theta,a\sin\theta\rangle$$ As can be checked because $||\hat n\,ds||=||d\vec r||$ and $\hat n\,ds\cdot d\vec r=0$ and $\hat n\,ds$ points out of the enclosed region. And $$\vec F=\langle\frac{a\cos\theta}{a^2\cos^2\theta+b^2\sin^2\theta},\frac{b\sin\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\rangle$$ Thus $$\int_{\Gamma}\vec F\cdot\hat n\,ds=\int_0^{2\pi}\frac{ab}{a^2\cos^2\theta+b^2\sin^2\theta}d\theta\tag1$$ Now, if $b=a$, then the path $\Gamma_1$ is a circle and the integral degenerates into $$\int_{\Gamma_1}\vec F\cdot\hat n\,ds=\int_0^{2\pi}d\theta=\left.\theta\right|_0^{2\pi}=2\pi$$ But in the area between the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and the circle $x^2+y^2=a^2$, we see that $$\vec\nabla\cdot\vec F=\frac{(1)(x^2+y^2)-x(2x)}{(x^2+y^2)^2}\frac{(1)(x^2+y^2)-y(2y)}{(x^2+y^2)^2}=0$$ So it follows by the divergence theorem that $$\int_{\Gamma}\vec F\cdot\hat n\,ds=\int_{\Gamma_1}\vec F\cdot\hat n\,ds=2\pi$$ This, along with eq. $(1)$ establishes the result.