How do integrate #int e^(cos)(t) (sin 2t) dt# between a = 0, #b = pi#?

Question

30043 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-10T09:37:09

Start with #int_0^pi e^cos(t)sin(2t) dt = 2int_0^pie^cos(t)sin(t)*cos(t)dt#

Now by part

#-2int-sin(t)e^cos(t)*cos(t) dt#

#u'=-sin(t)e^cos(t)#
#u=e^cos(t)#

#v = cos(t)#
#v' = -sin(t)#

You have : #-2([cos(t)*e^cos(t)]_0^pi - int_0^pi-sin(t)e^cos(t) dt)#

#=-2[cos(t)*e^cos(t)-e^cos(t)]_0^pi#

#=4/e#