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I would appreciate some assistance finding alternative approaches to solving the integral posed in the link below.

https://www.desmos.com/calculator/rstvj2v3cs

It describes and graphs a progression for decomposing the integral

$\int_0^{2\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx$

I am hoping for some insight into a simplified approach.

Much appreciated,

** Thanks to all for the feedback. I have completed the extended solution in desmos and welcome any additional feedback: https://www.desmos.com/calculator/g2lj84ofdu

Zachary Blakley
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  • At first made an image of your integrand. – Dr. Sonnhard Graubner Jun 13 '19 at 12:55
  • @Dr.SonnhardGraubner: There is an image, just click on the link... – Hans Lundmark Jun 13 '19 at 12:56
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    You have added a lot of details about your own attempt in the Desmos link (which is good). You should probably add that here as well, so people don't have to click links to see what your ideas are so far. – Mark Kamsma Jun 13 '19 at 12:57
  • That absolute value in there must go. Take the part of the integral from $\pi$ to $2\pi$, write $|\cos x|$ as $-\cos(x)$, and change variables to get $0$ to $\pi$. Add the two integrals. You get an integral with no $x$ in there. Easy to do. – GEdgar Jun 13 '19 at 13:37

2 Answers2

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It can be done in a simple way. Split the integral into four parts and substitute $$x=y,\qquad x={\pi\over2}+y,\qquad x=\pi+y,\qquad x={3\pi\over 2}+y$$ in the four parts. This leads to the integrals $$\eqalign{&\int_0^{\pi/2}{y\sin y\over2+\cos y}\>dy,\qquad \int_0^{\pi/2}{(\pi/2+y)\cos y\over2+\sin y}\>dy,\cr &\int_0^{\pi/2}{(\pi+y)(-\sin y)\over2+\cos y}\>dy,\qquad\int_0^{\pi/2}{(3\pi/2+y)(-\cos) y\over2+\sin y}\>dy .\cr} $$ The integrals with the same denominator can be collected, whereby the disagreeable parts disappear, and we are left with $$-\pi\int_0^{\pi/2}{\cos y\over 2+\sin y}\>dy-\pi\int_0^{\pi/2}{\sin y\over 2+\cos y}\>dy=2\pi\log{2\over3}=-2.5476\ .$$

Christian Blatter
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  • Thank you for the great insight into the problem. The four integral parts fit nicely with the visual I was hoping to create. I composed a final version of the (perhaps overly verbose) solution in desmos. If you have any additional feedback, it would be appreciated: link – Zachary Blakley Jun 17 '19 at 07:53
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\begin{align} &\int_0^{2\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx \\[1em] &=\int_0^{\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx + \int_{\pi}^{2\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx \end{align}

In the second integral, put $\pi+u=x$. This yields \begin{align} &\int_0^{\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx + \int_0^{\pi}\frac{(u+\pi)(-\sin u)}{2+\left|\cos u\right|}du \\[1em] &=\int_0^{\pi}\frac{x\sin x}{2+\left|\cos x\right|}+\frac{(x+\pi)(-\sin x)}{2+\left|\cos x\right|}dx \\[1em] &=-\pi \int_0^{\pi}\frac{\sin x}{2+\left|\cos x\right|} \end{align}

You should be able to solve it from here by putting $\cos x=u$.

I'm getting the final answer as $2\pi \ln \left(\frac{2}{3}\right)=−2.54761241$.

Lutz Lehmann
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Amit Rajaraman
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  • Thank you for the solution, I selected another answer as it fit more closely with the approach I was taking to generate a graphical split of the functions without the absolute value. If you are interested, I completed the [desmos solution] (https://www.desmos.com/calculator/g2lj84ofdu) – Zachary Blakley Jun 17 '19 at 07:55