How do you find the integral of #[x^4(sinx) dx] #?

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Answer 1 · 2018-06-13T12:39:05

#int x^4 sinx dx = (4x^3-24x) sinx - (x^4 -12 x^2 +24) cosx +C#

You need to integrate by parts repeatedly, to reduce the degree of #x#:

#int x^4 sinx dx = int x^4 d/dx (-cosx) dx#

#int x^4 sinx dx = -x^4cosx + 4int x^3 cosx dx#

#int x^4 sinx dx = -x^4cosx + 4int x^3 d/dx (sinx) dx#

#int x^4 sinx dx = -x^4cosx +4x^3sinx -12 int x^2 sinxdx#

#int x^4 sinx dx = -x^4cosx +4x^3sinx +12 x^2 cosx - 24 int x cosxdx#

#int x^4 sinx dx = -x^4cosx +4x^3sinx +12 x^2 cosx - 24x sinx +24 int sinxdx#

#int x^4 sinx dx = -x^4cosx +4x^3sinx +12 x^2 cosx - 24x sinx -24 cosx +C#