How to find integral of this function.
1 view (last 30 days)
Show older comments
Akshay Pratap Singh
on 2 Nov 2019
Answered: Akshay Pratap Singh
on 4 Nov 2019
clc
clear all
syms y1
k213 =(6858012277439187/(288230376151711744*cos(30*(3/2 - (3*y1)/2)^(1/2) - 10)*((-(8899397708189757*sin(30*(3/2 - (3*y1)/2)^(1/2) + 20))/(9007199254740992*cos(30*(3/2 - (3*y1)/2)^(1/2) - 10)))^(1/2) - 1)^2) - 1)*((3*(3/2 - (3*y1)/2)^(1/2))/2 - 1/2) + 1
k=int(k213, 0.5, 1)
2 Comments
Walter Roberson
on 2 Nov 2019
A closed form solution seems unlikely. Perhaps you should use vpaintegral()
Accepted Answer
John D'Errico
on 2 Nov 2019
Edited: John D'Errico
on 2 Nov 2019
No, that is not going to have any closed form solution.
One suggestion in the comments was to try vpaintegral. If it does not exist for you, then you either need to upgrade your version of MATLAB, or convert this to a double precision problem instead, then use integral. But integral will fail to yield anything useful, I predict. Even if it returns a number, that number will surely be random numerical garbage.
Anyway, even vpaintegral has issues with that kernel.
vpaintegral(k213,.5,1)
Error using sym/vpaintegral (line 202)
Failed precision goal. Try using 'MaxFunctionCalls'.
So lets look at what you have.
fun = matlabFunction(k213);
fplot(fun,[.5,1])
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/246139/image.jpeg)
When I finally stop laughing at the idea of using a numerical integration to integrate that, I'll just wish you good luck. Perhaps you have a friend, Don Quixote, who may have more success at the effort. He always was good at tilting at windmills.
The point is, all of those singularities are going to drive any numerical integration insane. And since a symbolic integral will surely not exist, perhaps you shoild consider if you really need to solve the problem at all. Not all problems you can pose have solutions.
If you really need to solve this, you might consider trying to identify the singularities. I see only 5 of them in the interval of interest. Isolate tham, then work on each one individually.
1 Comment
Steven Lord
on 2 Nov 2019
Another possibility, if you know that the integral of this function exists (if the integral has some sort of physical meaning) is to check that you've accurately implemented the function. pretty print the symbolic function and compare it to the mathematical function that you think you've implemented. Make sure you haven't divided when you should have multiplied or something similar.
pretty(vpa(k213, 6))
Or write that code in the Live Editor and display k213.
More Answers (1)
See Also
Categories
Find more on Calculus in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!