1. ## integration problem

in (sin(x) cos(x)) dx
let u=cos(x)
du=-sin(x) dx

if let u=sin(x)
du=cos(x) dx

if we put it into an right angle triangle length 3,4,5

then (a) get -0.18
but (b) get 0.32

2. ## Re: integration problem

Originally Posted by oponion
in (sin(x) cos(x)) dx
let u=cos(x)
du=-sin(x) dx

if let u=sin(x)
du=cos(x) dx

if we put it into an right angle triangle length 3,4,5

then (a) get -0.18
but (b) get 0.32
You have forgotten about the arbitrary constant that is added to each integral. Both answers are correct up to a different constant, since \displaystyle \begin{align*} -\frac{1}{2}\cos^2{x} = -\frac{1}{2}\left(1 - \sin^2{x}\right) = -\frac{1}{2} + \frac{1}{2}\sin^2{x} \end{align*}.

3. ## Re: integration problem

thank you very much
because my answer is different from the solution
and i use x=1,2 sub into my answer and the solution and get different answer
now i understand
thank you