x^2 cos 3x dx?

Question

Integration by parts - Step by step

Answer 1

Well Robert , your questions are solved by same way.

i just give the correct answer;
∫x^2 cos 3x dx = (2x Cos[3x])/9 + ((-2 + 9(x^2))Sin[3x])/27 + c

Good Luck...

Answer 2

This must be done in two steps. First set u=x^2, dv = cos3x; then v=sin3x/3

integ(u*dv) = uv - integ(v*du) => x^2*sin3x/3 - integ(2/3*x*sin3x*dx)

The second integral is also evaulated by parts; u=x, x, dv=sin3x, then v=-cos3x/3

x^2*sinx-2/3[-x*cos3x/3 - integ(-cos3x/3*dx)]

x^2*sin3x+2/3[x*cos3x/3-sin3x/3]

Answer 3

let dv=cos3x and x^2=u
integral =uv-int vdu
=x^2(1/3)sin3x-int(1/3)sin3x
(2x)dx
=x^2(1/3)sin3x-int2/3 xsin3x..................(1)
again for int xsin3x
taking sin3x as dv and x as du
xsin3x=-(1/3)xcos3x-int(1/3)sin3xdx
=-1/3 xcos3x+1/9cos3x+C
now substituting this in (1)
int x^2cos3x=(1/3)x^2sin(3x)
+(2/9)xcos(3x)-(2/27)cos(3x)+C

Answer 4

Do your own homework.