How do you integrate $\int {cos (7 x)}^{3} d x$ $intcos(7x)^3 dx$ ?

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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,

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

How do you integrate intcos(7x)^3 dx∫cos(7x)3dxintcos(7x)^3 dx?