How do you integrate #int xcosx# by parts from #[0,pi/2]#? Calculus Techniques of Integration Integration by Parts 1 Answer sjc Nov 21, 2016 #int_0^(pi/2)xcosxdx=pi/2-1# Explanation: Parts formula. #intu(dv)/(dx)dx=uv-intv(du)/(dx)dx# #int_0^(pi/2)xcosxdx# #u=x=>(du)/(dx)=1# #(dv)/(dx)=cosx=>v=sinx# #I=[xsinx-int(sinx)dx]_0^(pi/2)# #I=[xsinx+cosx]_0^(pi/2)# now substitute the limits in #I=[(pi/2)sin(pi/2)+cos(pi/2)]-[0xxsin0+cos0]# #I=[(pi/2)cancel(sin(pi/2))^1+cancel(cos(pi/2))^0]-[cancel(0xxsin0)^0+cancel(cos0)^1]# #int_0^(pi/2)xcosxdx=pi/2-1# 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 16568 views around the world You can reuse this answer Creative Commons License