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Question: Evaluate the following definite integral: $$ I= \int_{0}^{\pi/2} \ln\left(1 + a\cos\left(x\right) \over 1 - a\cos\left(x\right)\right)\, {{\rm d}x \over \cos\left(x\right)}\qquad \mbox{where}\qquad\left\vert\,a\,\right\vert < 1 $$

This is left as an exercise in my textbook. The book says that I should use the method of "differentiating parameters" by using the provided Leibniz Formula (?):

Prerequisite: Function $f(x,y)$ and its partial derivative $f_x(x,y)$ are both continuous on region $R=[a,b]\times[c,d]$. Function $\alpha(x)$ and $\beta(x)$ are both differentiable on intervl $[a,b]$ while $$ c \le \alpha(x) \le d,\quad c \le \beta(x) \le d \quad (a\le x\le b),$$

Conclusion: The Function $\displaystyle\Phi(x)=\int_{\alpha(x)}^{\beta(x)} f(x,y) \,\mathrm d y$ is differentiable on interval $[a,b]$, and $$\begin{align} \Phi^\prime(x) &= \frac{\mathrm d}{\mathrm d x} \int_{\alpha(x)}^{\beta(x)} f(x,y) \,\mathrm d y \\ &= \int_{\alpha(x)}^{\beta(x)} f_x(x,y) \,\mathrm d y + f\left[x,\beta(x)\right]\beta^\prime(x)-f\left[x,\alpha(x)\right]\alpha^\prime(x). \end{align}$$

First, I'd like to ask: What is the correct name of this theorem?

Next, here's my (failed?) attempts at solving the question.

Following one of the textbook examples on the same topic, I tried:

$$ \begin{align} I&=\int_0^\frac{\pi}{2} \left[\ln(1+y)\right]_{-a \cos x}^{a \cos x} \frac{\mathrm d x}{\cos x} \\ \\ &= \int_0^\frac{\pi}{2} \int_{-a \cos x}^{a \cos x} \frac{\mathrm d y}{1+y} \frac{\mathrm d x}{\cos x}. \\ &=\int_{-a}^{a} \frac{1}{1+y} \int_{0}^{\arccos\left|y/a\right|}\,\mathrm d x \mathrm d y \\ &=\int_{-a}^{a} \left.\left[\ln\left|\tan x + \sec x\right|\right]\right|_0^{\arccos(y/a)} \end{align} $$

And I don't know what to do next.

Also I tried to substitute $u=a \cos x$ but I don't know how to proceed.

Please help me with this (non-homework) problem by giving hints or solution, It's been a week since I first tried this problem. Thanks for helping!

P.S.: My MathJax doesn't render (or it may need hours of time), and I don't get preview for writing questions/answers. I'm relying on LaTeX now. How can I fix this problem? It used to work. I'm using IE 9.

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    $\begingroup$ Look a the Wikipedia article about differentiation under the integral sign. It has a useful worked example that yields to the same idea as your problem. Express your log as a difference of logs, differentiate with respect to $a$. That cancels the $\cos x$ at the bottom. $\endgroup$ – André Nicolas Oct 8 '12 at 5:41
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    $\begingroup$ LaTeX isn't my strong suit, but maybe if you have \begin{align} you don't want the dollar signs? Anyway, in the title, I think you want d, not \diff. $\endgroup$ – Gerry Myerson Oct 8 '12 at 6:19
  • $\begingroup$ I'll try to read all replies and corrected terms as early as possible. It's a tough day today and I'm worn out. I tried to add dollar signs around begin align because that's the way it worked in mathjax. LaTeX thinks \begin{align} to be $$. $\endgroup$ – Frenzy Li Oct 8 '12 at 19:20
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This is so called Leibniz integral rule. Using it you can easily compute this integral $$ \frac{d I}{d a} =\frac{d}{da}\int\limits_{0}^{\pi/2}\ln\frac{1+a\cos x}{1-a\cos x}\frac{dx}{\cos x} =\int\limits_{0}^{\pi/2}\frac{d}{da}\ln\frac{1+a\cos x}{1-a\cos x}\frac{dx}{\cos x} =\int\limits_{0}^{\pi/2}\frac{2dx}{1-a^2\cos^2 x} $$ $$ \frac{d I}{d a}=\int\limits_{0}^{\pi/2}\frac{2d(\tan x)}{(1-a^2)+\tan^2 x} =\int\limits_{0}^{+\infty}\frac{2dt}{(1-a^2)+t^2} $$ $$ \frac{d I}{d a}=2\frac{1}{\sqrt{1-a^2}}\arctan\frac{t}{\sqrt{1-a^2}}\Biggl|_0^\infty =\frac{\pi}{\sqrt{1-a^2}} $$ Since $I(0)=0$ we get $$ I(\alpha)=I(0)+\int\limits_0^{\alpha}\frac{dI}{da}da =\pi\int\limits_0^{\alpha}\frac{1}{\sqrt{1-a^2}}da =\pi\arcsin a\Biggl|_0^\alpha =\pi\arcsin\alpha $$

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  • $\begingroup$ Gee, this is brilliant, but how did you know we should differentiate it with respect to $a$? Is it being that you could cancel the $\cos x$ below $\mathrm{d} x$? What if the question says nothing about this Leibniz integral rule? $\endgroup$ – Frenzy Li Oct 9 '12 at 14:28
  • $\begingroup$ I also wonder why Mathematica fails to get the answer to this integral even after 1 hr of computing. I guess it needs to implement Leibniz integral rule? $\endgroup$ – Frenzy Li Oct 9 '12 at 14:40
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    $\begingroup$ When I see a hardcore integral I'm trying to apply Leibniz integral rule. Mathmetica won't implement this method because it might give wrong results. For exmple consider equality $$\int\limits_{0}^\infty\frac{\sin(ax)}{x}=\frac{\pi}{2}\mathrm{sign}(a)$$ $\endgroup$ – Norbert Oct 9 '12 at 15:18
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    $\begingroup$ @Norbert $$ {{\rm d} \over {\rm d}a}\int_{0}^{\infty} {\sin\left(\,ax\,\right) \over x}\,{\rm d}x = \pi\,\delta\left(\,a\,\right) $$ $\endgroup$ – Felix Marin Aug 31 '14 at 21:58
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    $\begingroup$ Please note that there is a small error in the computation: on the first line, the last integral should be multiplied by $2$, thus eliminating the $2$ in the denominator of the final result. $\endgroup$ – Alex M. Jun 9 '15 at 17:26

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