Ok so I'm supposed to evaluate:

$$\int_0^{\pi}\frac{x}{1-\sin{x}\cos{x}}\,\mathrm dx$$ I tried using definite integral properties, but this didn't yield any good follow up, because I couldn't evaluate the resulting integrals.

How exactly can one go about solving this? (preferably without contour integration)

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    $\begingroup$ Have you tried contour integration ? $\endgroup$ – user399481 Dec 23 '16 at 1:34
  • $\begingroup$ I haven't. I'm not too familiar with it. However, this is a level 3 problem for IIT, which is sn engineering entrance exam. So I'm sure that won't be necessary. $\endgroup$ – Kugelblitz Dec 23 '16 at 1:37
  • $\begingroup$ Can you try using the fact that $\sin x\cos x = \frac{\sin(2x)}{2}$? (Product-to-sum formula) $\endgroup$ – Calconym Dec 23 '16 at 1:38
  • $\begingroup$ @Kugelblitz, So you haven't had a class of complex variables ? $\endgroup$ – user399481 Dec 23 '16 at 1:41
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    $\begingroup$ I agree, this problem is hard to solve with normal methods, but straightforward with contour integration. Perhaps you are expected to know some. $\endgroup$ – Kaynex Dec 23 '16 at 1:47

$$\int_{0}^{\pi} \frac{2x}{2-\sin 2x}\text{d}x=\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{2x}{2-\sin 2x}\text{d}x}_{I_1}+\underbrace{\int_{\frac{\pi}{2}}^{\pi} \frac{2x}{2-\sin 2x}\text{d}x}_{I_2}$$ Solving for $I_1$ and $I_2$ involves some trig identity knowledge, more specifically $\sin \left (\frac{\pi}{2}-x\right )=\cos x$ and $\sin (k\pi-x)=\sin x$ if $k$ is odd. Making the substitution $x=\frac{\pi}{2}-u$ this changes the integral into $\int_{0}^{\frac{\pi}{2}} \frac{\pi-2u}{2-\sin 2u}\text{d}u$ after switching the bounds and taking care of the negative. Now set $u=x$ and rewrite the integral in terms of $x$, add them together to get $$2I_1=\int_{0}^{\frac{\pi}{2}} \frac{\pi}{2-\sin 2x}\text{d}x\implies I_1=\frac{\pi}{2}\int\frac{1}{2-\sin 2x}\text{d}x$$ Making the Weierstrass substitution $t=\tan x$ we obtain \begin{align*} I_1&=\frac{\pi}{2} \int_{0}^{\infty} \frac{1}{(1+t^2)\left ( 2-\frac{2t}{1+t^2} \right )}\text{d}t\\ &=\frac{\pi}{4}\int_{0}^{\infty}\frac{1}{\left ( t-\frac{1}{2} \right )^2+\left (\frac{\sqrt 3}{2} \right )^2}\text{d}t\\ &=\frac{\pi}{2\sqrt 3}\tan^{-1}\left (\frac{2t-1}{\sqrt 3}\right )\biggr\rvert_{0}^{\infty}\\ &=\frac{\pi^2}{3\sqrt 3} \end{align*} doing something extremely similar to $I_2$ but with the substitution $x=\frac{3\pi}{2}-u$ we get that $I_2=\frac{\pi^2}{2\sqrt 3}$ adding the two integrals we get the final answer of $$\boxed{\frac{5\pi^2}{6\sqrt 3}}$$

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    $\begingroup$ Nice answer... I did this way abt 30 mins ago during lunch. W-sub is a hack I tell you... $\endgroup$ – Kugelblitz Dec 23 '16 at 7:59
  • $\begingroup$ @Kugelblitz I know what you mean. When I first heard of it I was like no way... I call it the miracle substitution because that's what it looks like to me haha $\endgroup$ – Teh Rod Dec 23 '16 at 8:00
  • $\begingroup$ And that matches with Harnoor Lal's deleted answer. $\endgroup$ – marty cohen Dec 23 '16 at 19:42

Note: I made a simple error (my specialty) in the changed limits of integration. I had $\pi/2$ where I should of had $2\pi$. I have made the appropriate changes.

Thanks to heropup for pointing out my mistake.

Here's a start.

$\begin{array}\\ \int_0^{\pi}\frac{x}{1-\sin{x}\cos{x}}dx &=\int_0^{\pi}\frac{x}{1-\sin(2x)/2}dx\\ &=\int_0^{\pi}\frac{2x}{2-\sin(2x)}dx\\ &=\frac12\int_0^{2\pi}\frac{x}{2-\sin(x)}dx\\ &=\frac14\int_0^{2\pi}\frac{x}{1-\sin(x)/2}dx\\ &=\frac14\int_0^{2\pi}x\, dx\sum_{n=0}^{\infty}(\sin(x)/2)^n\\ &=\frac14\sum_{n=0}^{\infty}\int_0^{2\pi}x\, dx(\sin(x)/2)^n\\ &=\frac14\sum_{n=0}^{\infty}\frac1{2^n}\int_0^{2\pi}x\sin^n(x) dx\\ \end{array} $

According to Wolfy, if $I_n =\int_0^{2\pi}x\sin^n(x) dx $, then, for $n=0, 1, 2, 3, 4,5,6,7,8,9$, $I_n = 2\pi^2, -2\pi, \pi^2,-4\pi/3, 3\pi^2/4,-16\pi/15, 5\pi^2/8,-32\pi/35, 35\pi^2/64,-256\pi/315 $.

From this, it seems pretty certain that $I_{2n} =a_n \pi^2 $ and $I_{2n+1} =-b_n\pi $ where all of $a_n, b_n$ are rational and appear almost guessible.

From these results, there should be a recurrence of $I_n$ in terms of $I_{n-2}$, possibly gotten by the standard trick of two integration by parts.

The integral $\int_0^{2\pi}x\cos^n(x) dx $ might come into this also.

Another possibility is to use $I_n =I_{n-2}-\int_0^{2\pi} x \cos^2(x)\sin^{n-2}(x)dx $, but I don't see where to go from this.

It's late, and I'm tired, so I'll leave it at this.

Here are the recurrences I mentioned, courtesy of Table of Integrals, Series, and Products, Seventh Edition, by I.S. Gradshteyn and I.M. Ryzhik:

From G&R, p. 214.

$\begin{array}\\ \int x^m \sin^n x\, dx &= \dfrac{x^{m−1} \sin^{n−1} x}{n^2} (m\sin x − nx \cos x)\\ & +\dfrac{n − 1}{n}\int x^m \sin^{n−2} xdx − \dfrac{m(m − 1)}{n^2} \int x^{m−2} \sin^n xdx\\ \text{if } m=1\\ \int x \sin^n x\, dx &= \dfrac{ \sin^{n−1} x}{n^2} (\sin x − nx \cos x)\\ & +\dfrac{n − 1}{n}\int x \sin^{n−2} xdx \\ \text{if } m=1, n=1\\ \int x \sin x\, dx &= (\sin x − x \cos x)\\ \text{if } m=1, n=0\\ \int x dx &= \dfrac{x^2}{2}\\ \text{if } m=1, n \ge 2\\ \int_0^{2\pi} x \sin^n x\, dx &= \dfrac{ \sin^{n−1} x}{n^2} (\sin x − nx \cos x)|_0^{2\pi}\\ & +\dfrac{n − 1}{n}\int_0^{2\pi} x \sin^{n−2} xdx \\ &= \dfrac{n − 1}{n}\int_0^{2\pi} x \sin^{n−2} xdx \\ \text{if } m=1, n=1\\ \int_0^{2\pi} x \sin x\, dx &= (\sin x − x \cos x)|_0^{2\pi}\\ &= -2\pi\\ \text{if } m=1, n=0\\ \int_0^{2\pi} x dx &= \dfrac{x^2}{2}|_0^{2\pi}\\ &=2\pi^2\\ \text{so that}\\ I_0 &= 2\pi^2\\ I_1 &=-2\pi\\ I_{2n} &= \dfrac{2n-1}{2n}I_{2n-2}\\ I_{2n+1} &= \dfrac{2n}{2n+1}I_{2n-1}\\ \end{array} $

I could use these to get explicit formulas for $I_{2n}$ and $I_{2n+1}$, but since Teh Rod has done such a nice job of getting the actual answer, I'll stop right here.

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    $\begingroup$ You've made an error in your substitution step; the upper limit should be $2\pi$, not $\pi/2$. $\endgroup$ – heropup Dec 23 '16 at 7:05
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    $\begingroup$ Thanks. Corrected (I hope) and upvoted. $\endgroup$ – marty cohen Dec 23 '16 at 7:26
  • $\begingroup$ Thanks a lot for the useful info! $\endgroup$ – Kugelblitz Dec 24 '16 at 0:56

$$ \large \int_0^\pi \frac{\text{d}x}{1-\sin{x}\cos{x}} \underset{\text{divide}}{\stackrel{\text{phase shift}}{=}} \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{\sec^2{x}\text{d}x}{\tan^2{x}-\tan{x}+1} \stackrel{t=\tan{x}}{=} \int_{-\infty}^\infty \frac{\text{d}t}{t^2-t+1}$$

And the rest is trivial.


$$I=\int_0^\pi\frac{x}{1-\sin(x)\cos(x)}\ dx=\int_0^{\pi}\frac{2x}{2-\sin(2x)}\ dx$$

$$=\frac12\int_0^{2\pi}\frac{x}{2-\sin(x)}\ dx\overset{x+\pi/2=t}{=}\frac12\int_{\pi/2}^{5\pi/2}\frac{t-\pi/2}{2+\cos(t)}\ dt$$

Using the identity

$$\frac{1}{a+b\cos(t)}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$

We have


$$\Longrightarrow I=\frac{1}{2\sqrt{3}}\underbrace{\int_{\pi/2}^{5\pi/2}\left(t-\frac{\pi}{2}\right)\ dt}_{2\pi^2}+\frac{1}{\sqrt{3}}\sum_{n=1}^\infty (\sqrt{3}-2)^n\underbrace{\int_{\pi/2}^{5\pi/2}\left(t-\frac{\pi}2\right)\cos(nt)\ dt}_{A}$$

$$A=\int_{\pi/2}^{5\pi/2}t\cos(nt)\ dt-\frac{\pi}{2}\underbrace{\int_{\pi/2}^{5\pi/2}\cos(nt)\ dt}_{0}=\frac{2\pi}{n}\sin\left(\frac{\pi}{2}n\right), \quad n=1,2,3,...$$

$$\Longrightarrow I=\frac{\pi^2}{\sqrt{3}}+\frac{2\pi}{\sqrt{3}}\underbrace{\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^n}{n}\sin\left(\frac{\pi}{2}n\right)}_{-\pi^2/12}=\frac{5\pi^2}{6\sqrt{3}}$$

Details for the last result;

$$\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^n}{n}\sin\left(\frac{\pi}{2}n\right)=\Im\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^ne^{in\pi/2}}{n}=\Im\sum_{n=1}^\infty \frac{[i(\sqrt{3}-2)]^n}{n}$$


  • $\begingroup$ That's pretty damn cool! $\endgroup$ – Kugelblitz Mar 30 '20 at 16:58

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