What is the antiderivative of # x*(sinx)^2#?

Question

Emily Ammerman · Accepted Answer

#1/4(x^2 -xsin2x -1/2 cos 2x) + C#

We must compute:

#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)

Presently, Our integral becomes

#int x ((1 - cos 2x)/2)dx#

#= int x/2 (1 - cos 2x) dx#

We are going to use Integration by Parts now.

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#

Thus, 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#

Consequently, The Complete Integration (From (iii)),

#= 1/4x^2 - 1/4xsin2x - 1/8cos 2x + C#

#= 1/4(x^2 - xsin2x - 1/2cos 2x) + C#

I hope this is helpful.

Scarlett Allison · Answer

undefined The antiderivative of $ x \cdot (\sin x)^2 $ with respect to $ x $ is:

$$ -\frac{1}{2} x \cos x + \frac{1}{4} \sin(2x) + C $$

Where $ C $ is the constant of integration.