How do integrate #int e^(cos)(t) (sin 2t) dt# between a = 0, #b = pi#? Calculus Techniques of Integration Integration by Parts 1 Answer Tom Apr 10, 2015 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# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 30167 views around the world You can reuse this answer Creative Commons License