The answer this trigonometric integration example is wrong despite attempts to remedy it. Anyone knows why?

[Stephen23]

https://www.mathworks.com/help/matlab/ref/inline.html:

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

Answer 1

[Ameer Hamza]

It is the correct result, it is just a different way of writing it.

>> syms x
>> f = 2*cos(x) + sin(x)
f =
2*cos(x) + sin(x)
>> int_f = int(f)
int_f =
-2*cos(x/2)*(cos(x/2) - 2*sin(x/2))
>> simplify(diff(int_f)) % same expression back
ans =
2*cos(x) + sin(x)

[Jan]

Thanks alot

[Ameer Hamza]

I am glad to be of help!!!

[Steven Lord]

If you want to test if Symbolic Math Toolbox can prove that two symbolic expressions that look different actually represent the same quantity you can use isAlways.

>> syms x
>> f1 = sin(x)
f1 =
sin(x)
 
>> f2 = (exp(1i*x)-exp(-1i*x))/2i
f2 =
(exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2
 
>> isAlways(f1 == f2)
ans =
  logical
   1

See Euler's formula.