Question

A integral de sin^3 (3x) dx?

Answer 1

$cos9x36-cos3x4+С$

Answer 2

$\int\limits {sin^3(3x)} \, dx$

utilizando a identidade trigonometrica
$sen^2(x) +cos^2(x) = 1\\\\ \boxed{\boxed{sen^2(x) = 1-cos^2(x)}}$

e como
$\boxed{\boxed{sen^3(3x)}}= sen^2(3x)*sen(3x) = \boxed{\boxed{[1-cos^2(3x)] * sen(3x)}}$

fazendo essa multiplicaçao temos
$[1-cos^2(3x)]] * sen(3x)=\\\\=\boxed{\boxed{sen(3x) -sen(3x)*cos^2(3x)}}$

substituindo na integral ficamos com
$\int\limits {sen(3x)-sen(3x)*cos^2(3x)} \, dx \\\\\\= \boxed{\boxed{\int\limits {sen(3x)} \, dx - \int\limits {sen(3x)*[cos(3x)]^2} \, dx }}$

resolvendo as integrais
$\int\limits {sen(3x)} \, dx$

substituindo 
$u = 3x\\\\du = 3..dx\\\\ \frac{du}{3}=dx$

integrando

$\int\limits {sen(u)} \, \frac{du}{3} = \frac{1}{3} \int\limits {sen(u)} \, du } = \frac{1}{3}*|-cos(u) |= \frac{-cos(3x)}{3}$

resolvendo a segunda integral
$\int\limits {sen(3x)*[cos(3x)]^2} \, dx$

aplicando a  substituiçao
$\boxed{u = cos(3x)}\\\\ \frac{du}{dx} = -sen(3x)*3 \\\\ \boxed{ \frac{du}{-3sen(3x)} =dx}$

temos
$\int {sen(3x)* u^2} \, \frac{du}{-3*sen(3x)} = \int {u^2} \frac{du}{-3} =\boxed{- \frac{1}{3} \int\limits {u^2.du} \, }$

calculando a integral
$\frac{-1}{3} | \frac{u^3}{3}|= \frac{-u^3}{9}$

como u = cos(3x)
$\frac{-1}{3} \int{u^3} \, du = \frac{-cos^3(3x)}{9}$

então a integral fica

$\int\limits {sen(3x)} \, dx - \int\limits {sen(3x)*[cos(3x)]^2} \, dx }\\\\ = \frac{-cos(3x) }{3} - [- \frac{cos^3(3x)}{9}] \\\\=\boxed{\boxed{ \frac{-cos(3x) }{3} + \frac{cos^3(3x)}{9} +C }}$


resposta
 
$\boxed{\boxed{ \int\limits {sin^3(3x)} \, dx = \frac{ -cos(3x)}{3} + \frac{cos^3(3x)}{9} +C}}$