Integration of #x^3cosx# ?

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Answer 1 · 2018-05-22T11:03:15

#int \ x^3 cos(x) \ dx = (3x^2-6)cos x+(x^3-6x) sin x+C#

The integral will take some form like:

#P(x)cos(x) + Q(x)sin(x) + C#

where #P(x)# and #Q(x)# are polynomials of degree #<= 3# and #C# is the constant of integration.

Then:

#d/(dx)(P(x)cos(x) + Q(x)sin(x) + C)#

#=(P'(x)+Q(x))cos(x)+(Q'(x)-P(x))sin(x)#

So equating coefficients, we want to solve:

#{ (P'(x) + Q(x) = x^3), (Q'(x)-P(x) = 0) :}#

Hence:

#Q''(x) + Q(x) = x^3#

Hence:

#Q(x) = x^3-6x#

and:

#P(x) = Q'(x) = 3x^2-6#

So:

#int \ x^3 cos(x) \ dx = (3x^2-6)cos x+(x^3-6x) sin x+C#