Question

integral using integration by parts
x^2 cos x dx

Answer 1

Compute I = ∫ x^2 cos x dx by parts.

Let u = x^2, dv = cos x dx. Then du = 2x dx, v = sin x. So
I = x^2 sin x − ∫ 2x sin x dx.

Evaluate J = ∫ 2x sin x dx by parts:
take u2 = 2x, dv2 = sin x dx ⇒ du2 = 2 dx, v2 = −cos x. Then
J = 2x(−cos x) − ∫ 2(−cos x) dx = −2x cos x + 2 ∫ cos x dx
= −2x cos x + 2 sin x.

Thus
I = x^2 sin x − (−2x cos x + 2 sin x) + C
= x^2 sin x + 2x cos x − 2 sin x + C.

(You can also write I = sin x (x^2 − 2) + 2x cos x + C.)