Which is the correct solution: -1/2*cos^2(x) or 1/2*sin^2(x)?

[echy5555]

Homework Statement

The integral of sin(x)*cos(x) either equals -1/2*cos^2(x) or 1/2*sin^2(x). Which is it? It can't be both, right?

Homework Equations

\intudv=u*v-\intvdu

The Attempt at a Solution

integration by parts:
u=cos(x) dv=sin(x)dx
du=-sin(x)dx v=-cos(x)

-cos^2(x)-\intsin(x)*cos(x)
so:
2*\intsin(x)*cos(x)=cos^2(x)

but can't it also be done this way?

u=sin(x) dv=cos(x)dx
du=cos(x)dx v=sin(x)

sin^2(x)-\intsin(x)*cos(x)
so:
2*\intsin(x)*cos(x)=sin^2(x)

 

[losiu99]

Actually, it can. Antiderivatives are unique only up to additive constant, and
\frac{1}{2}\sin^2 x = \frac{1}{2}(1-\cos^2 x)=\frac{1}{2} - \frac{1}{2}\cos^2 x
and so, derivatives of this two functions are equal.

 

[echy5555]

Okay, I get it. If you put the limits of integration into the integral then you get the same answer, but if you leave it as an indefinite integral, the constants of integration are different. Right?

 

[xaos]

the constant is arbitrary, so a fixed constant plus an arbitrary constant is still arbitrary. think of indefinite integration as generating a stack of vertically identical antiderivatives two of which are the same minus some constant from each. but what happens when you find the total indefinite integration of a multiple integral?

 

[echy5555]

It all makes sense now. Thank you.

To answer your question,

When you find the total indefinite integration of a multiple integral, then the "constants of integration" are not actually constants, but function of the other independent variable.