6
$\begingroup$

Writing:

Integrate[ArcTan[(a Cos[x] + b Sin[x])^2], {x, 0, 2 Pi}, Assumptions -> a^2 + b^2 > 0]

$$\int_0^{2\pi}\arctan\left((a\cos x + b\sin x)^2\right)dx,$$ where $a $ and $b $ are real numbers.

I get:

2 Pi ArcTan[Sqrt[1/2 (-1 + Sqrt[1 + (a^2 + b^2)^2])]]

$$2\pi\arctan\sqrt{\frac{\sqrt{1 + (a^2 + b^2)^2}-1}2} $$

How to derive this result on paper?

$\endgroup$

migrated from mathematica.stackexchange.com Jun 16 at 19:05

This question came from our site for users of Wolfram Mathematica.

  • 2
    $\begingroup$ I have translated your question into math language. Please check. $\endgroup$ – user Jun 16 at 19:36
7
$\begingroup$

Denote $$C := a^2 + b^2 .$$ Then, we can find an angle $x_0$ such that $a = \sqrt{C} \cos x_0$ and $b = -\sqrt{C} \sin x_0$. The angle sum formula for $\sin$ lets us rewrite the quantity in the inner parentheses of the integrand as $$a \sin x + b \cos x = \sqrt{C} \sin (x - x_0).$$ Then, appealing to the periodicity of the integrand lets us rewrite the integral as $$I(C) := \int_0^{2 \pi} \arctan (C \sin^2 x) \,dx .$$ Differentiating under the integral sign gives $$I'(C) = \int_0^{2 \pi} \frac{\sin^2 x\, dx}{1 + C^2 \sin^4 x} .$$ Now, use symmetry to rewrite $I'(C)$ in terms of an integral over $[0, \pi]$, and apply the Euler substitution $x = 2 \arctan t, \,dx = \frac{2\,dt}{1 + t^2}$, giving the rational integral $$I'(C) = 16 \int_0^{\infty} \frac{t^2 (t^2 + 1) \,dt}{(t^2 + 1)^4 + 16 C^4 t^4}.$$ Integrating gives $$I'(C) = \frac{\pi \sqrt{2}}{\sqrt{1 + \sqrt{1 + C^2}} \sqrt{1 + C^2}} .$$ (This can be done with contour integration, which in this case is tedious but straightforward. Quite possibly there is a better method, and I would be grateful to learn it.) Since $I(0) = 0$, we have $$I(C) = \pi \sqrt{2} \int_0^C \frac{dc}{\sqrt{1 + \sqrt{1 + c^2}} \sqrt{1 + c^2}} = 2 \pi \sqrt{2} \int_0^{u_0} \frac{du}{u^2 + 2} ;$$ the latter equality follows from applying the substitution $c^2 + 1 = (u^2 + 1)^2$, and $u_0$ is the $u$-value corresponding to $c = C$. The integral on the right-hand side is elementary, and so one can produce an explicit formula for $I(C)$ in terms of $C$ and hence in terms of $a, b$.

$\endgroup$
  • $\begingroup$ Fabulous, thanks! $\endgroup$ – TeM Jun 17 at 6:05
7
$\begingroup$
  1. We have $a\cos x+b\sin x=r\cos(x-\phi)$, where $$r=\sqrt{a^2+b^2},\quad\cos\phi=a/r,\quad\sin\phi=b/r,$$ and we can simply replace $x-\phi$ by $x$ in the integrand (because of its $2\pi$-periodicity). Denoting $c=(a^2+b^2)/2$, we see that the given integral is equal to $$\int_{0}^{2\pi}\arctan(2c\cos^2 x)\,dx=\int_{0}^{2\pi}\arctan\big(c(1+\cos x)\big)\,dx.$$

  2. Recall that if $d\in\mathbb{C}$ such that $|d|<1$, then $$\int_{0}^{2\pi}\ln(1-2d\cos x+d^2)\,dx=0$$ (assuming the principal branch taken). This can be seen, after $$1-2d\cos x+d^2=(1-de^{ix})(1-de^{-ix}),$$ as an application of the Cauchy integral theorem. (Alternatively, one can use the above and power series, or even just split $\int_{0}^{2\pi}=\int_{0}^{\pi}+\int_{\pi}^{2\pi}$ and substitute $x=y+\pi$ in the second integral, to get $I(d)=I(d^2)/2$ from which $I(d)=0$ follows easily.) This implies $$\int_{0}^{2\pi}\ln(1+d\cos x)\,dx=2\pi\ln\frac{1+\sqrt{1-d^2}}{2}.$$

  3. Write $$\arctan\big(c(1+\cos x)\big)=\frac{1}{2i}\ln\frac{1+ic}{1-ic}\frac{1+d\cos x}{1+\bar{d}\cos x},\qquad d=\frac{ic}{1+ic}$$ (where $\bar{d}$ is complex conjugate to $d$); the integral then equals $$2\pi\arg(1+ic+\sqrt{1+2ic})=2\pi\arctan\frac{c+v}{1+u}=2\pi\arctan v,$$ where $\sqrt{1+2ic}=u+iv$, and we use $u=c/v$. This is the answer.

$\endgroup$
  • $\begingroup$ Thank you very much to you, even if it is hard to understand, I don't know the field of complex numbers well. $\endgroup$ – TeM Jun 17 at 6:05
4
$\begingroup$

Premise: this self-response has been added "taking inspiration" from the two previous ones, especially that of @Travis.


Since, as usual, it is possible to write: $$ \begin{aligned} A\cos x+B\sin x & = \sqrt{A^2 + B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\cos x + \frac{B}{\sqrt{A^2+B^2}}\sin x\right) \\ & = C\left(\cos\varphi\cos x + \sin\varphi\sin x\right) \\ & = C\cos(x-\varphi) \end{aligned} $$ it follows that: $$ I(C) := \int_0^{2\pi}\arctan\left(\left(A\cos x+B\sin x\right)^2\right)\text{d}x = \int_0^{2\pi}\arctan\left(\left(C\cos(x-\varphi)\right)^2\right)\text{d}x $$ that placing $y = x - \varphi$, thanks to the periodicity of the integrand, equivalent to writing (1): $$ I(C) = \int_{0-\varphi}^{2\pi-\varphi}\arctan\left(\left(C\cos y\right)^2\right)\text{d}y = \int_0^{2\pi}\arctan\left(C^2\cos^2 y\right)\text{d}y\,. $$ Then, differentiating under the integral sign, we have: $$ \begin{aligned} I'(C) & = \int_0^{2\pi} \frac{2C\cos^2 y}{1+C^4\cos^4 y}\,\text{d}y \\ & = \int_0^{\frac{\pi}{2}} \frac{8C\frac{1+\cos(2y)}{2}}{1+C^4\left(\frac{1+\cos(2y)}{2}\right)^2}\,\text{d}y \\ & = \int_0^{\pi} \frac{4C\left(1+\cos(z)\right)}{1+\frac{C^4}{4}\left(1+\cos(z)\right)^2}\,\frac{1}{2}\text{d}z \\ & = \int_0^{\infty} \frac{2C\left(1+\frac{1-t^2}{1+t^2}\right)}{1+\frac{C^4}{4}\left(1+\frac{1-t^2}{1+t^2}\right)^2}\,\frac{2}{1+t^2}\text{d}t \\ & = \int_0^{\infty} \frac{8C}{C^4+\left(1+t^2\right)^2}\text{d}t \\ & = \int_0^{\infty} \frac{8C}{\left(C^2 + \text{i}\left(1+t^2\right)\right)\left(C^2 - \text{i}\left(1+t^2\right)\right)}\text{d}t \\ & = \frac{4\text{i}}{C}\left(\int_0^{\infty} \frac{1}{1+\text{i}C^2+t^2}\text{d}t - \int_0^{\infty} \frac{1}{1-\text{i}C^2+t^2}\text{d}t\right) \\ & = \frac{4\text{i}}{C}\left(\frac{1}{\sqrt{1+\text{i}C^2}}\int_0^{\infty} \frac{\frac{1}{\sqrt{1+\text{i}C^2}}}{1+\left(\frac{t}{\sqrt{1+\text{i}C^2}}\right)^2}\text{d}t - \frac{1}{\sqrt{1-\text{i}C^2}}\int_0^{\infty} \frac{\frac{1}{\sqrt{1-\text{i}C^2}}}{1+\left(\frac{t}{\sqrt{1-\text{i}C^2}}\right)^2}\text{d}t\right) \\ & = \frac{4\text{i}}{C}\left(\frac{\arctan\left(\frac{t}{\sqrt{1+\text{i}C^2}}\right)}{\sqrt{1+\text{i}C^2}}-\frac{\arctan\left(\frac{t}{\sqrt{1-\text{i}C^2}}\right)}{\sqrt{1-\text{i}C^2}}\right)_{t=0}^{t=\infty} \\ & = \frac{4\text{i}}{C}\left(\frac{\frac{\pi}{2}}{\sqrt{1+\text{i}C^2}}-\frac{\frac{\pi}{2}}{\sqrt{1-\text{i}C^2}}\right) - \frac{4\text{i}}{C}\left(\frac{0}{\sqrt{1+\text{i}C^2}}-\frac{0}{\sqrt{1-\text{i}C^2}}\right) \\ & = \frac{2\pi}{C}\frac{\text{i}\left(\sqrt{1-\text{i}C^2}-\sqrt{1+\text{i}C^2}\right)}{\sqrt{1+C^4}} \\ \end{aligned} $$ and noting that: $$ \left(\frac{\text{i}\left(\sqrt{1-\text{i}C^2}-\sqrt{1+\text{i}C^2}\right)}{\sqrt{1+C^4}}\right)^4 = \left(\frac{-2\left(-1+\sqrt{1+C^4}\right)}{1+C^4}\right)^2 $$ we get: $$ I'(C) = \frac{2\pi}{C}\sqrt{\frac{2\left(-1+\sqrt{1+C^4}\right)}{1+C^4}}\,, $$ while, integrating, we have (2): $$ \begin{aligned} I(C) & = 2\pi\int \frac{2}{C\sqrt{1+C^4}}\sqrt{\frac{-1+\sqrt{1+C^4}}{2}}\,\text{d}C \\ & = 2\pi\int \frac{\frac{2}{C\sqrt{1+C^4}}\sqrt{\frac{-1+\sqrt{1+C^4}}{2}}\left(1+\frac{-1+\sqrt{1+C^4}}{2}\right)}{1+\frac{-1+\sqrt{1+C^4}}{2}}\,\text{d}C \\ & = 2\pi\int \frac{\frac{2}{C\sqrt{1+C^4}}\,\frac{C^4}{4\sqrt{\frac{-1+\sqrt{1+C^4}}{2}}}}{1+\frac{-1+\sqrt{1+C^4}}{2}}\,\text{d}C \\ & = 2\pi\int \frac{\frac{1}{2\sqrt{\frac{-1+\sqrt{1+C^4}}{2}}}\,\frac{C^3}{\sqrt{1+C^4}}}{1+\frac{-1+\sqrt{1+C^4}}{2}}\,\text{d}C \\ & = 2\pi\arctan\left(\sqrt{\frac{-1+\sqrt{1+C^4}}{2}}\,\right) + k\,. \end{aligned} $$ Given that from (1) we have $I(0) = 0$ and from (2) we have $I(0) = k$, we deduce that $k = 0$ and therefore: $$ \int_0^{2\pi}\arctan\left(C^2\cos^2 y\right)\text{d}y = 2\pi\arctan\left(\sqrt{\frac{-1+\sqrt{1+C^4}}{2}}\,\right) $$ ie: $$ \int_0^{2\pi}\arctan\left(\left(A\cos x+B\sin x\right)^2\right)\text{d}x = 2\pi\arctan\left(\sqrt{\frac{-1+\sqrt{1+\left(A^2+B^2\right)^2}}{2}}\,\right), $$ as we wanted to prove.

$\endgroup$

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.