Question

How do you integrate `(x^2)*Sin[x^(3/2)]dx`?

Answer 1

`= 2/3 ( sin x^(3/2) - x^(3/2) cos x^(3/2) ) + C`

Explanation:

firstly note that `d/dx cos x^(3/2) = -3/2 x^(1/2) sin x^(3/2`

So
`I = int \ x^2 \ sinx^(3/2) \ dx`

can be written as

`I = -2/3 int \ x^(3/2) \d/dx ( cos x^(3/2) )\ dx`

which by IBP gives

`-3/2 I = x^(3/2) cos x^(3/2)- int \d/dx ( x^(3/2) ) cos x^(3/2) \ dx`

`-3/2 I = x^(3/2) cos x^(3/2)- int \3/2 x^(1/2) cos x^(3/2) \ dx`

noting that `d/dx sin(x^(3/2)) = 3/2x^(1/2) cos x^(3/2`

we have `-3/2 I = x^(3/2) cos x^(3/2)- int \ d/dx sin x^(3/2) \ dx`

`-3/2 I = x^(3/2) cos x^(3/2)- sin x^(3/2) + C`

`I = -2/3 ( x^(3/2) cos x^(3/2)- sin x^(3/2)) + C`

`= 2/3 ( sin x^(3/2) - x^(3/2) cos x^(3/2) ) + C`

Here a sub may have been better

`I(x) = int \ x^2 \ sinx^(3/2) \ dx`

`u = x^(3/2), x = u^(2/3)`

So `du = 3/2 x^(1/2) dx = 3/2 u^(1/3) dx`

So the integration becomes

`I(u) = int \u^(4/3) \ sin u * \3/2 u^(- 1/3)\ du`

`I(u) =3/2 int \u \ sin u \ \ du`

But you still need IBP

`2/3 I(u) = int \ u d/(du) (- cos u) \ du`

`2/3 I(u) = - u cos u - int \d/(du) u (- cos u) \ du`

`2/3 I(u) = - u cos u + int \ cos u \ du`

`I(u) =3/2( - u cos u + sin u) + C`

so

`I(x) = 3/2( sin x^(3/2) - x^(3/2) cos x^(3/2) ) + C`

Not really.