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

Question

32687 views around the world

You can reuse this answer
Creative Commons License

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 $x$ :

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

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

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

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

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

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

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

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