We have to calculate:-
intx(sin x)^2dx∫x(sinx)2dx.
= int xsin^2xdx=∫xsin2xdx
As, We know, cos 2x = 1 - 2sin^2xcos2x=1−2sin2x,
Then 2sin^2x = 1 - cos 2x2sin2x=1−cos2x
rArr sin^2 x = (1 - cos 2x)/2⇒sin2x=1−cos2x2....................(i)
Now, Our integral Turns into
int x ((1 - cos 2x)/2)dx∫x(1−cos2x2)dx
= int x/2 (1 - cos 2x) dx=∫x2(1−cos2x)dx
Now We will use Integration by Parts.
So, int x/2 (1 - cos 2x)dx∫x2(1−cos2x)dx
= x/2 int (1 - cos 2x)dx - int(d/dx(x/2)int (1 - cos 2x))dx=x2∫(1−cos2x)dx−∫(ddx(x2)∫(1−cos2x))dx
= x/2 (int dx - int cos2xdx) - int(1/2(intdx - int cos 2x dx)dx=x2(∫dx−∫cos2xdx)−∫(12(∫dx−∫cos2xdx)dx............(ii)
Now, Lets solve int cos2xdx∫cos2xdx.
Substitute u = 2xu=2x.
So, du = 2dx rArr dx = 1/2 dudu=2dx⇒dx=12du
So, int cos2xdx∫cos2xdx
= 1/2intcos u du=12∫cosudu
= 1/2sinu + C=12sinu+C
= 1/2sin2x + C=12sin2x+C
So, From (i),
We have,
= x/4 (2x - sin2x) - 1/4int(2x - sin2x)dx=x4(2x−sin2x)−14∫(2x−sin2x)dx
= x/4(2x - sin2x) - 1/4 xx 2intxdx + 1/4int sin 2xdx=x4(2x−sin2x)−14×2∫xdx+14∫sin2xdx
= x/4(2x - sin2x) - 1/4x^2 + 1/4int sin 2xdx=x4(2x−sin2x)−14x2+14∫sin2xdx
= 1/2x^2 - 1/4xsin2x - 1/4x^2 + 1/4intsin2xdx=12x2−14xsin2x−14x2+14∫sin2xdx.................(iii)
Now,
intsin2xdx∫sin2xdx
Substitute u = 2x rArr du = 2dx rArr dx = 1/2duu=2x⇒du=2dx⇒dx=12du
So,
intsin2xdx∫sin2xdx
= 1/2intsinudu=12∫sinudu
= -1/2cosu + C=−12cosu+C
= -1/2cos 2x + C=−12cos2x+C
So, The Final Integration (From (iii)),
= 1/4x^2 - 1/4xsin2x - 1/8cos 2x + C=14x2−14xsin2x−18cos2x+C
= 1/4(x^2 - xsin2x - 1/2cos 2x) + C=14(x2−xsin2x−12cos2x)+C
Hop ethis helps.