#x^2 cosx dx# intregrate?

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2 Answers
paho
May 10, 2018

#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#.

Binayaka C.
May 10, 2018

#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]

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