Plz solve integration problem

[monu2010]

i am integrating it by two methods

a=integration{(sin(x).cos(x).dx}

i).first using by parts
a=int{sin(x).cos(x)}
I II
a=[sin(x).sin(x)] - int{cos(x).sin(x)}]
a=(sin(x))^2 - a
2a=(sin(x)^2)
a=(sin(x)^2)/2
put x=30
a=.125 --------------------------------------------------------(1)

ii).by using trigonometry relation
a=int{sin(x).cos(x)}
a=(1/2).int{2.sin(x).cos(x)}
a=(1/2)int{sin(2x)}
a=(1/2).(-cos(2x))/2

put x=30
a= -0.125 ----------------------------------------------------(2)
how is it possible

help!!!!!

[Glaysher]

Corrections made

i am integrating it by two methods

a=integration{(sin(x).cos(x).dx}

i).first using by parts
a=int{sin(x).cos(x)}
I II
u = sin (x) and v' = cos(x)
u' = cos(x) and v = - sin(x)

a= - [sin(x).sin(x)] + int{cos(x).sin(x)}]
a=-(sin(x))^2 + a

Not useful

ii).by using trigonometry relation
a=int{sin(x).cos(x)}
a=(1/2).int{2.sin(x).cos(x)}
a=(1/2)int{sin(2x)}
a=(1/2).(-cos(2x))/2

put x=30
a= -0.125 ----------------------------------------------------(2)
how is it possible

Correct

help!!!!!

[Mattywoo2]

Yes, unfortunately you missed a negative, so your first solution isn't quite right.

Glaysher is on fire today!!