Is there an easier way to solve this complicated integral?

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In summary, the conversation discusses different methods of integrating the given integral, which involves constants and an infinite series. The solution obtained from using Maple and Mathematica is complex and not easily simplified. A suggestion is made to substitute and use integration by parts to possibly simplify the integral.
  • #1
gaganaut
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While I was working on my research problem, I came across this integral.

[tex]\int_0^{\pi}\;\frac{k_1}{-k_2\left(2R\;(k_3\cos{\alpha}+k_4\sin{\alpha})-k_5\right)^{1.5}}[/tex]

All the [tex]k_i\;'s[/tex] are constants and so is R.

I tried writing out an infinite series for [tex]\left(2R\;(k_3\cos{\alpha}+k_4\sin{\alpha})-k_5\right)^{-1.5}[/tex] in Maple and then integrating. But the solution is nasty with the first 4 terms considered. It has got quite a few terms to deal with.

I then tried performing numerical integration by using the Simpson's rule, but the solution again has too many terms in it.

So is there any easier way to solve this or will I have to live with the multi-term solution that I get from both the methods I used?

Any sort of help will be highly appreciated.

Thanks
 
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  • #2
I did the integration in Mathematica and it gave me
[tex]-\frac{2 F_1\left(-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{2};\left\{\frac{2 R c_3+c_5}{2 R \sqrt{\frac{c_3^2}{c_4^2}+1} c_4+c_5},\frac{c_5-2 R c_3}{2 R
\sqrt{\frac{c_3^2}{c_4^2}+1} c_4+c_5}\right\},\left\{-\frac{2 R c_3+c_5}{2 R \sqrt{\frac{c_3^2}{c_4^2}+1} c_4-c_5},\frac{2 R c_3-c_5}{2 R
\sqrt{\frac{c_3^2}{c_4^2}+1} c_4-c_5}\right\}\right) c_1 \left(\sqrt{2 R c_3-c_5} \sqrt{\frac{R \left(c_3^2+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_3+c_4^2\right)}{2
R \left(c_3^2+c_4^2\right)-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}} \sqrt{\frac{R \left(c_3^2-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_3+c_4^2\right)}{2 R
\left(c_3^2+c_4^2\right)+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}}+\sqrt{-2 R c_3-c_5} \sqrt{\frac{R \left(c_3^2-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4
c_3+c_4^2\right)}{2 R \left(c_3^2+c_4^2\right)-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}} \sqrt{\frac{R \left(c_3^2+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4
c_3+c_4^2\right)}{2 R \left(c_3^2+c_4^2\right)+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}}\right)}{R c_2 c_4 \sqrt{-2 R c_3-c_5} \sqrt{2 R c_3-c_5}}[/tex]

where F1 is some Appell hypergeometric function...
so somehow I'm not convinced that there is a "nice" solution.
 
  • #3
Well its straight forward to get it into the form

[tex]\int_{0}^{\pi}\frac{Adx}{(sin(x+\alpha)-B)^\frac{3}{2}}[/tex]

where A, B and alpha are functions of R and the K's

you could then try a substitution like

[tex]y=sin(x+\alpha)+A\;\;\;\;;dx=\frac{dy}{cos(arcsin(y-A)))}[/tex]

to give

[tex]\int_{sin(\alpha)+A}^{sin(\alpha+\pi)+A}\frac{Ady}{sin(arcsin(y-A)-\frac{\pi}{2})x^\frac{3}{2}}[/tex]

maybe then try integration by parts...
 

Related to Is there an easier way to solve this complicated integral?

1. What is a complicated integral?

A complicated integral is an expression that involves the sum or difference of infinitely many terms, where each term is a product of a function and a differential. It is considered complicated because it requires advanced mathematical techniques to solve.

2. Why are complicated integrals important?

Complicated integrals are important in many areas of science and engineering, including physics, chemistry, and economics. They are used to calculate quantities such as area, volume, mass, and probability, and are essential for solving complex real-world problems.

3. How do you solve a complicated integral?

Solving a complicated integral requires a combination of techniques, such as substitution, integration by parts, and trigonometric identities. It also involves knowledge of advanced concepts in calculus, such as the Fundamental Theorem of Calculus and the Chain Rule.

4. Can complicated integrals be solved analytically?

While some complicated integrals can be solved analytically (i.e. by finding a closed-form solution), many cannot. In these cases, numerical methods such as the trapezoidal rule or Simpson's rule can be used to approximate the value of the integral.

5. Are there any real-world applications of complicated integrals?

Yes, complicated integrals have numerous real-world applications, such as determining the center of mass of an object, calculating the work done by a force, and finding the probability of events in statistics. They are also used in the development of mathematical models for physics, economics, and other fields.

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