bobandjohn
10-05-2007, 01:33 AM
∫ e^x sinx cosx dx
= ∫ e^x * (1/2) sin(2x) dx
= (1/2) ∫ sin(2x) d(e^x)
= (1/2) [sin(2x)*e^x - ∫ e^x * 2cos(2x) dx ]
= (1/2) [sin(2x)*e^x - 2 ∫ cos(2x) d(e^x)]
= (1/2) sin(2x)*e^x - [e^x*cos(2x) - ∫ e^x (-2sin(2x))dx]
= (1/2) sin(2x)*e^x - e^x*cos(2x) - 4∫e^x * sinx * cosx dx
= (1/2) sin(2x)*e^x - e^x*cos(2x) - 4I
so
5I = (1/2) sin(2x)*e^x - e^x*cos(2x)
5I = e^x/2 [sin(2x) - 2cox(2x)]
I = e^x/10 [sin(2x) - 2cox(2x)]
is there anything wrong with the working any mis multi or divide? thanks
= ∫ e^x * (1/2) sin(2x) dx
= (1/2) ∫ sin(2x) d(e^x)
= (1/2) [sin(2x)*e^x - ∫ e^x * 2cos(2x) dx ]
= (1/2) [sin(2x)*e^x - 2 ∫ cos(2x) d(e^x)]
= (1/2) sin(2x)*e^x - [e^x*cos(2x) - ∫ e^x (-2sin(2x))dx]
= (1/2) sin(2x)*e^x - e^x*cos(2x) - 4∫e^x * sinx * cosx dx
= (1/2) sin(2x)*e^x - e^x*cos(2x) - 4I
so
5I = (1/2) sin(2x)*e^x - e^x*cos(2x)
5I = e^x/2 [sin(2x) - 2cox(2x)]
I = e^x/10 [sin(2x) - 2cox(2x)]
is there anything wrong with the working any mis multi or divide? thanks