Thread: integration problem

in (sin(x) cos(x)) dx
let u=cos(x)
du=-sin(x) dx
answer is -1/2cos^2(x) (a)

if let u=sin(x)
du=cos(x) dx
then answer is 1/2*sin^2(x) (b)

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

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

Originally Posted by oponion

in (sin(x) cos(x)) dx
let u=cos(x)
du=-sin(x) dx
answer is -1/2cos^2(x) (a)

if let u=sin(x)
du=cos(x) dx
then answer is 1/2*sin^2(x) (b)

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 .

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