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

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!

Answer 1

[Dimitris Kalogiros]

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.

[Dimitris Kalogiros]

You can verify the relation between what you got and what you expect, using the following trigonometric formula:

[Steven Lord]

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]

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!