Evaluate the integral?

Question

How do I do this problem? please show me step by step. Thanks.

evaluate the integral

∫ x^2 cos mxdx

Answer 1

∫ x^2 cos mxdx

Integrate by parts

let u = x² ; du = 2xdx

dv = cos(mx)dx ; v = ∫cos(mx)dx = sin(mx)/m + C

∫ x² cos mxdx  = x² sin(mx)/m - 2/m∫x sin(mx)dx + C

Integrate by parts

∫x sin(mx)dx

u =x ; du = dx

dv = sin(mx)dx ; v = ∫sin(mx)dx = -cos(mx)/m + C

∫x sin(mx)dx  = -x cos(mx)/m + 1/m∫cos(mx)dx

∫x sin(mx)dx  = -x cos(mx)/m + sin(mx)/m² + C

∫ x² cos mxdx  = x² sin(mx)/m - 2/m[-x cos(mx)/m + sin(mx)/m²] + C

∫ x² cos mxdx  = x² sin(mx)/m + 2x cos(mx)/m² - 2sin(mx)/m³ + C

Answer 2

Use integration by parts to reduce the powers of x² by differentiation until you get cosine by itself:

∫ x²·cos( m·x ) dx

Let u = x²

Then du = 2·x dx

Let dv = cos( m·x ) dx

Then v = sin( m·x )/m

→ u·v - ∫ v du

= [ x² ] · [ sin( m·x )/m ]  - ∫ [ sin( m·x )/m ] · [ 2·x dx ]

= x²·sin( m·x )/m - ∫ 2·x·sin( m·x )/m dx

Let q = 2·x

Then dq = 2 dx

Let dr = sin( m·x )/m dx

Then r = -cos( m·x )/m²

→  x²·sin( m·x )/m - { q·r - ∫ r dq }

= x²·sin( m·x )/m - { [ 2·x ] · [ -cos( m·x )/m² ] - ∫ [ -cos( m·x )/m² ] [ 2 dx ] }

= x²·sin( m·x )/m + 2·x·cos( m·x )/m² - ∫ 2·cos( m·x )/m² dx

= x²·sin( m·x )/m + 2·x·cos( m·x )/m² - 2·sin( m·x )/m³ + C