How to calculate indefinite integral?

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Don Donatas
Don Donatas el 3 de Mzo. de 2015
Respondida: Sean de Wolski el 4 de Mzo. de 2015
Hello, I hope you den help me. I am doing my math homework and I want to check the answer in Matlab. However, I checked that the answer, that is given by Matlab, is wrong. Here is an example of what I wrote:
>> syms x >> int(sin(2*x-3), x) >> ans = sin(x-3/2)^2
But the correct answer is -cos(3-2x)/2 + C So maybe you can tell me what I do wrong because I am new in Matlab so it is quite hard. Thank you very much for your help!
Best regards, Don

Respuestas (2)

Brendan Hamm
Brendan Hamm el 3 de Mzo. de 2015
Editada: Brendan Hamm el 4 de Mzo. de 2015
First off the answer is -cos(2*x-3)/2 + C. The two statements are equivalent up to a constant. Using the half angle identity:
sin(u)^2 = 1/2 - 1/2*cos(2*u)
let u = x -3/2
sin(x-3/2)^2 + C = 1/2 - 1/2*cos(2*(x-3/2)) + C
sin(x-3/2)^2 + C = -1/2*cos(2*x-3) + D

Sean de Wolski
Sean de Wolski el 4 de Mzo. de 2015
Both answers are right, just simplified differently:
syms x
mupad_answer = int(sin(2*x-3), x)
your_answer = -cos(3-2*x)/2;
isequal(simplify(diff(your_answer)),simplify(diff(mupad_answer)))

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