How do u solve the integral below ?

#int_ #x sin^3 x dx

1 Answer
Mar 12, 2018

#I=1/36[-27xcosx+27sinx+3xcos3x-sin3x]+c#

Explanation:

We know that,
#color(red)(sin3A=3sinA-4sin^3A=>4sin^3A=3sinA-sin3A)#
#color(red)(=>sin^3A=1/4[3sinA-sin3A])#
So,
#I=intxsin^3xdx=1/4[intx(3sinx-sin3x)dx]##=1/4int3xsinxdx-color(blue)(1/4intxsin3xdx)#
Using,#color(red)(int(u*v)dx=uintvdx=int(u^'*intvdx)dx)#
#I=1/4[(3x)(-cosx)-int(3)(-cosx)dx]-color(blue)(1/4[x((-cos3x)/3)-int(1)((-cos3x)/3)dx])#
#I=-3/4xcosx+3/4sinx+color(blue)(1/12xcos3x-1/12(sin3x)/3+c#
#I=1/36[-27xcosx+27sinx+3xcos3x-sin3x]+c#