How do you integrate #intcos(7x)^3 dx#?

Question

2051 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T09:44:56

The answer is #=-1/(42i^(1/6))Gamma(1/3","-343ix^3)-1/(42i^(1/6))Gamma(1/3","343ix^3)+C#

We need

#costheta=(e^(itheta)+e^(-itheta))/2#

#i^2=-1#

Therefore,

#intcos(7x)^3dx=intcos(343x^3)dx#

#=int((e^(i343x^3)+e^(-i343x^3))/2)dx#

#=1/2inte^(i343x^3)dx+1/2inte^(-i343x^3)dx#

Let #u=-7i^(1/6)x#, #=>#, #du=-7i^(1/6)dx#

#1/2inte^(i343x^3)dx=-1/(14i^(1/6))inte^(-u^3)du#

#=-1/(42i^(1/6))Gamma (1/3 "," u^3)#

#=-1/(42i^(1/6))Gamma(1/3","-343ix^3)#

And

#1/2inte^(-i343x^3)dx=-1/(14i^(1/6))inte^(u^3)du#

#=-1/(42i^(1/6))Gamma(1/3","-u^3)#

#=-1/(42i^(1/6))Gamma(1/3","343ix^3)#

Finally,

#intcos(7x)^3dx=-1/(42i^(1/6))Gamma(1/3","-343ix^3)-1/(42i^(1/6))Gamma(1/3","343ix^3)+C#