How do you find the indefinite integral of #int x(cos(8x))^2dx#?

Question

1910 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-17T14:25:12

#I = x^2/4+1/32xsin16x+1/512cos16x+C#

#I=int x(cos8x)^2dx#

Integratioin By Parts:

#int udv = uv - int vdu#

#u=x => du=dx#

#(cos8x)^2=cos^2 8x = (1+cos16x)/2#

#dv= (cos8x)^2dx#

#v=int (cos8x)^2dx = 1/2 int (1+cos16x)dx#

#v=1/2(x+1/16sin16x)#

#I = x/2(x+1/16sin16x) - 1/2int (x+1/16sin16x)dx#

#I = x/2(x+1/16sin16x) - 1/2(x^2/2-1/16*1/16cos16x)+C#

#I = 1/2(x^2+1/16xsin16x-x^2/2+1/256cos16x)+C#

#I = x^2/4+1/32xsin16x+1/512cos16x+C#