Why do I get "sin(t)^2" when I "int(sin(2*t))"?

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Peter Hammond
Peter Hammond 2018-9-17
评论： Peter Hammond 2018-9-17
I enter the following at the start of a Matlab session:
t = sym('t'); y = sin(2*t); int(y)
Matlab returns the following: ans = sin(t)^2
This is clearly not correct. It should return -cos(2*t)/2. Any other number in the cosine argument besides "2" seems to give the correct answer (e.g. if y = sin(3*t), int(y) returns -cos(3*t)/3). I've tried this in v2014b as well as v2016a. Simulink does the same thing if you integrate the function sin(2*t). Either this is a bug, or I'm going insane!

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Dimitris Kalogiros
Dimitris Kalogiros 2018-9-17
编辑：Dimitris Kalogiros 2018-9-17
Results of int(), are correct. Actually sin(t)^2 and -cos(2t)/2 differ only by a constant 1/2.
And this is legal, since at every integral we can consider a constant that is added to the result.
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Steven Lord
Steven Lord 2018-9-17
The NIST Handbook of Mathematical Functions has a section with trigonometric function identities. On this page, equation 4.21.28 states that cos(2*x) = 1-2*sin^2(x) So through a little manipulation of that identity you can see what Dimitris said is true, the expression you expected and the expression Symbolic Math Toolbox provided differ only by a constant.
So this is not a bug and you're not going insane. The packaging looks different, but the contents inside the package are the same.
Peter Hammond
Peter Hammond 2018-9-17
Thank you both! I get it. The offset created by squaring the sine expression threw me off since the negative part of the wave disappeared. It helps to be familiar with the trig identities!

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