Question

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

Answer 1

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

Explanation:

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.