Integral of sin(x)*cos(x)

Question

anonymous · Answer

u = cos x and 
du = -sin x

anonymous · Answer

but if i change 
u = sin x
du = cos x

you will get 1/2 sin^2 x but why isn't this right?

anonymous · Answer

the answer is -1/2 cos^2 x

anonymous · Answer

let sinx=t
then cosx dx=dt

$$\int\limits_{?}^{?}tdt  =t^2=c=\sin^2x$$

turingtest · Answer

u=sinx    du=cosx$$\int\limits udu=u^2/2+C=(\sin^2x)/2+C$$ or u=cos x du=-sinx$$\int\limits -udu=-u^2/2+C=(-\cos^2x)/2+C$$ I guess they are the same...

anonymous · Answer

the strange this is that they aren't the same :O

turingtest · Answer

oh because sin^2x=1-cos^x=-cos^2+c
so this happens because it is indefinite. the constant C is 1

anonymous · Answer

oehhh you lost me there with your sin^2x=1-cos^x=-cos^2+c

turingtest · Answer

As long as they only differ by a constant is's ok to get different answers. This happens a lot in more advanced intregals that require integration by parts.
$$\sin^2x+\cos^2x=1$$$$\sin^2x=1-\cos^2$$
which, because 1 is a constant, we can write as
$$\sin^2=C-\cos^2x=-\cos^x+C$$
where here C=1

turingtest · Answer

Once again, this is also because this is an indefinite integral and so requires a "+C". If it were a definite integral you would probably get into complications of some sort here.

anonymous · Answer

hmm... but i can use the 0.5sin^2 x?

or is this just wrong? because everyone uses the -0.5cos^2 x

turingtest · Answer

I don't see why not... 
take the derivative of 0.5sin^2(x)+C=sin(x)cos(x)
so 0.5sin^2(x) is clearly an anti-derivative of sin(x)cos(x).

anonymous · Answer

i thought so too! hmm.. weird that wolfram alpha also gives the answer of -0.5cos^2 x and not the sin x version

but thank you for your explanation

turingtest · Answer

yeah I just looked at wolfram as well, seems weird. Hopefully your teacher can answer, or if you see JamesJ on this site ask him, he's one of the best.

anonymous · Answer

oke will do :D thanks :)

anonymous · Answer

<3