Question

`x^2 cosx dx` intregrate?

Answer 1

`int x^2 cos x\ d x = ( x^2 - 2 ) sin x + 2 x cos x + C`

Explanation:

This integral can be solved by using integration by parts twice:

`int x^2 cos x\ d x =`

`= x^2 sin x - 2 int x sin x\ d x =`

`= x^2 sin x - 2 [ - x cos x + int cos x\ d x ] =`

`= x^2 sin x + 2 x cos x - 2 sin x + C =`

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

Answer 2

`int x^2 cos x dx =x^2 sin x + 2 x cos x - 2 sin x + c`

Explanation:

`int x^2 cos x dx` let `u= x^2:. du = 2x dx`

`dv = cosx dx or int dv = int cos x dx or v = sin x`

`int x^2 cos x dx = uv - int v du = x^2 sin x - int sinx *2x dx` or

`int x^2 cos x dx =x^2 sin x - 2int sinx *x dx`

let `u= x:. du = dx ,dv = sin x dx or v = -cos x`

`int sinx *x dx = - x cos x - int -cosx dx` or

`int sinx *x dx = - x cos x + int cosx dx` or

`int sinx *x dx = - x cos x + sinx`

`int x^2 cos x dx =x^2 sin x - 2(- x cos x + sin x )` or

`int x^2 cos x dx =x^2 sin x + 2 x cos x - 2 sin x + c` [Ans]