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

Mar 9, 2018

#### Answer:

The answer is $= - \frac{1}{42 {i}^{\frac{1}{6}}} \Gamma \left(\frac{1}{3} \text{,"-343ix^3)-1/(42i^(1/6))Gamma(1/3",} 343 i {x}^{3}\right) + C$

#### Explanation:

We need

$\cos \theta = \frac{{e}^{i \theta} + {e}^{- i \theta}}{2}$

${i}^{2} = - 1$

Therefore,

$\int \cos {\left(7 x\right)}^{3} \mathrm{dx} = \int \cos \left(343 {x}^{3}\right) \mathrm{dx}$

$= \int \left(\frac{{e}^{i 343 {x}^{3}} + {e}^{- i 343 {x}^{3}}}{2}\right) \mathrm{dx}$

$= \frac{1}{2} \int {e}^{i 343 {x}^{3}} \mathrm{dx} + \frac{1}{2} \int {e}^{- i 343 {x}^{3}} \mathrm{dx}$

Let $u = - 7 {i}^{\frac{1}{6}} x$, $\implies$, $\mathrm{du} = - 7 {i}^{\frac{1}{6}} \mathrm{dx}$

$\frac{1}{2} \int {e}^{i 343 {x}^{3}} \mathrm{dx} = - \frac{1}{14 {i}^{\frac{1}{6}}} \int {e}^{- {u}^{3}} \mathrm{du}$

$= - \frac{1}{42 {i}^{\frac{1}{6}}} \Gamma \left(\frac{1}{3} \text{,} {u}^{3}\right)$

$= - \frac{1}{42 {i}^{\frac{1}{6}}} \Gamma \left(\frac{1}{3} \text{,} - 343 i {x}^{3}\right)$

And

$\frac{1}{2} \int {e}^{- i 343 {x}^{3}} \mathrm{dx} = - \frac{1}{14 {i}^{\frac{1}{6}}} \int {e}^{{u}^{3}} \mathrm{du}$

$= - \frac{1}{42 {i}^{\frac{1}{6}}} \Gamma \left(\frac{1}{3} \text{,} - {u}^{3}\right)$

$= - \frac{1}{42 {i}^{\frac{1}{6}}} \Gamma \left(\frac{1}{3} \text{,} 343 i {x}^{3}\right)$

Finally,

$\int \cos {\left(7 x\right)}^{3} \mathrm{dx} = - \frac{1}{42 {i}^{\frac{1}{6}}} \Gamma \left(\frac{1}{3} \text{,"-343ix^3)-1/(42i^(1/6))Gamma(1/3",} 343 i {x}^{3}\right) + C$