We have to calculate:-
#intx(sin x)^2dx#.
#= int xsin^2xdx#
As, We know, #cos 2x = 1 - 2sin^2x#,
Then #2sin^2x = 1 - cos 2x#
#rArr sin^2 x = (1 - cos 2x)/2#....................(i)
Now, Our integral Turns into
#int x ((1 - cos 2x)/2)dx#
#= int x/2 (1 - cos 2x) dx#
Now We will use Integration by Parts.
So, #int x/2 (1 - cos 2x)dx#
#= x/2 int (1 - cos 2x)dx - int(d/dx(x/2)int (1 - cos 2x))dx#
#= x/2 (int dx - int cos2xdx) - int(1/2(intdx - int cos 2x dx)dx#............(ii)
Now, Lets solve #int cos2xdx#.
Substitute #u = 2x#.
So, #du = 2dx rArr dx = 1/2 du#
So, #int cos2xdx#
#= 1/2intcos u du#
#= 1/2sinu + C#
#= 1/2sin2x + C#
So, From (i),
We have,
#= x/4 (2x - sin2x) - 1/4int(2x - sin2x)dx#
#= x/4(2x - sin2x) - 1/4 xx 2intxdx + 1/4int sin 2xdx#
#= x/4(2x - sin2x) - 1/4x^2 + 1/4int sin 2xdx#
#= 1/2x^2 - 1/4xsin2x - 1/4x^2 + 1/4intsin2xdx#.................(iii)
Now,
#intsin2xdx#
Substitute #u = 2x rArr du = 2dx rArr dx = 1/2du#
So,
#intsin2xdx#
#= 1/2intsinudu#
#= -1/2cosu + C#
#= -1/2cos 2x + C#
So, The Final Integration (From (iii)),
#= 1/4x^2 - 1/4xsin2x - 1/8cos 2x + C#
#= 1/4(x^2 - xsin2x - 1/2cos 2x) + C#
Hop ethis helps.