# Why is my integration output the mathematical expression and not the solution?

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Ayush Kumar on 13 Mar 2023
Commented: Star Strider on 13 Mar 2023
I am trying to solve the following question:
and I have utlised the following script to achieve so:
syms x
func = ((x^3*cos(x/2)+1/2)*(sqrt(4-x^2)));
int(func,x,-2,2)
But MATLAB keeps returning the explicit form of the function back instead of computing it, as shown below:
ans =
int((4 - x^2)^(1/2)*(x^3*cos(x/2) + 1/2), x, -2, 2)
What am I doing wrong that is making it not compute but merely display the function?
The answer should be or π.
Walter Roberson on 13 Mar 2023
The output IBP is the same as the input. At the limits +/- 2 the function is 0.
Star Strider on 13 Mar 2023
Running integrateByParts correctly this time (that I had not done previously, proving once again that paying close attention to the documentation is always appropriate) produces —
syms x
func = ((x^3*cos(x/2)+1/2)*(sqrt(4-x^2)));
F = int(func,x,-2,2)
iBP1 = integrateByParts(F,diff(sqrt(4-x^2)))
Fi1 = release(iBP1)
Fi1 = simplify(Fi1, 500)
iBP2 = integrateByParts(F,diff((x^3*cos(x/2)+1/2)))
Fi2 = release(iBP2)
Fi2 = simplify(Fi2, 500)
F =
int((4 - x^2)^(1/2)*(x^3*cos(x/2) + 1/2), x, -2, 2)
iBP1 =
-int((4 - x^2)^(1/2)*(2*x^3*cos(x/2) - ((x^2 - 4)*(x^3*cos(x/2) + 1/2))/x^2 + ((x^2 - 4)*(3*x^2*cos(x/2) - (x^3*sin(x/2))/2))/x + 1), x, -2, 2)
Fi1 =
-int((4 - x^2)^(1/2)*(2*x^3*cos(x/2) - ((x^2 - 4)*(x^3*cos(x/2) + 1/2))/x^2 + ((x^2 - 4)*(3*x^2*cos(x/2) - (x^3*sin(x/2))/2))/x + 1), x, -2, 2)
Fi1 =
-Inf
iBP2 =
-int(x^3*cos(x/2)*((4 - x^2)^(1/2) + ((4 - x^2)^(1/2)*(x^3*cos(x/2) + 1/2)*((x^3*cos(x/2))/4 - 6*x*cos(x/2) + 3*x^2*sin(x/2)))/(3*x^2*cos(x/2) - (x^3*sin(x/2))/2)^2 - (x*(x^3*cos(x/2) + 1/2))/((4 - x^2)^(1/2)*(3*x^2*cos(x/2) - (x^3*sin(x/2))/2))), x, -2, 2)
Fi2 =
-int(x^3*cos(x/2)*((4 - x^2)^(1/2) + ((4 - x^2)^(1/2)*(x^3*cos(x/2) + 1/2)*((x^3*cos(x/2))/4 - 6*x*cos(x/2) + 3*x^2*sin(x/2)))/(3*x^2*cos(x/2) - (x^3*sin(x/2))/2)^2 - (x*(x^3*cos(x/2) + 1/2))/((4 - x^2)^(1/2)*(3*x^2*cos(x/2) - (x^3*sin(x/2))/2))), x, -2, 2)
Fi2 =
NaN
I ran this in MATLAB Online, since it requires more than the allotted 55 seconds to run it here. It turns out that different definitions of (as described in the documentation) produce different results, neither of them finite.
.

VBBV on 13 Mar 2023
func = @(x) ((x.^3.*cos(x./2)+1/2).*(sqrt(4-x.^2)));
integral(func,-2,2)
If you use integral it will work as you expected
VBBV on 13 Mar 2023
Edited: VBBV on 13 Mar 2023

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