Question

Integral por Substituição Trigonométrica:
em anexo
preciso do desenvolvimento e que feche com a resposta

Answer 1

Comecemos pela nossa integral original que desejamos calcular:

$I_{0}=\int{\cos^{6}3x\,dx}$


Para avaliar esta integral, aplicaremos repetidamente o método de integração por partes. Para o $k-$ésimo passo de integração por partes, sempre teremos que

$\int{u_{k}\,dv_{k}}=u_{k}v_{k}-\int{v_{k}\,du_{k}}$


$\bullet\;\;k=0:$

Reescrevendo $I_{0}$:

$I_{0}=\int{\cos^{5}3x\cdot \cos 3x\,dx}\\ \\ \\ \begin{array}{ll} u_{0}=\cos^{5}3x\;\;&\;\;du_{0}=-15\cos^{4}3x\cdot \mathrm{sen\,}3x\,dx\\ \\ dv_{0}=\cos 3x\,dx\;\;&\;\;v_{0}=\dfrac{1}{3}\mathrm{\,sen\,}3x \end{array}\\ \\ \\ \\ \int{u_{0}\,dv_{0}}=u_{0}v_{0}-\int{v_{0}\,du_{0}}\\ \\ \\ I_{0}=\dfrac{1}{3}\cos^{5}3x\cdot \mathrm{sen\,}3x+5\int{\cos^{4}3x\cdot\mathrm{sen^{2}\,}3x\,dx} \\ \\ \\ I_{0}=\dfrac{1}{3}\cos^{5}3x\cdot \mathrm{sen\,}3x+5\int{\cos^{4}3x\cdot(1-\cos^{2}3x )\,dx}\\ \\ \\ I_{0}=\dfrac{1}{3}\cos^{5}3x\cdot \mathrm{sen\,}3x+5\int{(\cos^{4}3x-\cos^{6}3x)\,dx}\\ \\ \\ I_{0}=\dfrac{1}{3}\cos^{5}3x\cdot \mathrm{sen\,}3x+5\int{\cos^{4}3x\,dx}-5\int{\cos^{6}3x\,dx}\\ \\ \\ I_{0}=\dfrac{1}{3}\cos^{5}3x\cdot \mathrm{sen\,}3x+5\int{\cos^{4}3x\,dx}-5I_{0}$


$I_{0}+5I_{0}=\dfrac{1}{3}\cos^{5}3x\cdot \mathrm{sen\,}3x+5\int{\cos^{4}3x\,dx}\\ \\ \\ 6I_{0}=\dfrac{1}{3}\cos^{5}3x\cdot \mathrm{sen\,}3x+5\int{\cos^{4}3x\,dx}\\ \\ \\ I_{0}=\dfrac{1}{18}\cos^{5}3x\cdot \mathrm{sen\,}3x+\dfrac{5}{6}\int{\cos^{4}3x\,dx}$


Fazendo $I_{1}=\int{\cos^{4}3x\,dx},$ temos

$I_{0}=\dfrac{1}{18}\cos^{5}3x\cdot \mathrm{sen\,}3x+\dfrac{5}{6}I_{1}$


$\bullet\;\;k=1:$

Reescrevendo $I_{1}$:

$I_{1}=\int{\cos^{3}3x\cdot \cos 3x\,dx}\\ \\ \\ \begin{array}{ll} u_{1}=\cos^{3}3x\;\;&\;\;du_{1}=-9\cos^{2} 3x\cdot \mathrm{sen\,}3x\,dx\\ \\ dv_{1}=\cos 3x\,dx\;\;&\;\;v_{1}=\dfrac{1}{3}\mathrm{\,sen\,}3x \end{array}\\ \\ \\ \\ \int{u_{1}\,dv_{1}}=u_{1}v_{1}-\int{v_{1}\,du_{1}}\\ \\ \\ I_{1}=\dfrac{1}{3}\cos^{3}3x\cdot \mathrm{sen\,}3x+3\int{\cos^{2}3x\cdot \mathrm{sen^{2}\,}3x\,dx}\\ \\ \\ I_{1}=\dfrac{1}{3}\cos^{3}3x\cdot \mathrm{sen\,}3x+3\int{\cos^{2}3x\cdot (1-\cos^{2}3x)\,dx}\\ \\ \\ I_{1}=\dfrac{1}{3}\cos^{3}3x\cdot \mathrm{sen\,}3x+3\int{(\cos^{2}3x-\cos^{4}3x)\,dx}\\ \\ \\ I_{1}=\dfrac{1}{3}\cos^{3}3x\cdot \mathrm{sen\,}3x+3\int{\cos^{2}3x\,dx}-3\int{\cos^{4}3x\,dx}\\ \\ \\ I_{1}=\dfrac{1}{3}\cos^{3}3x\cdot \mathrm{sen\,}3x+3\int{\cos^{2}3x\,dx}-3I_{1}$


$I_{1}+3I_{1}=\dfrac{1}{3}\cos^{3}3x\cdot \mathrm{sen\,}3x+3\int{\cos^{2}3x\,dx}\\ \\ \\ 4I_{1}=\dfrac{1}{3}\cos^{3}3x\cdot \mathrm{sen\,}3x+3\int{\cos^{2}3x\,dx}\\ \\ \\ I_{1}=\dfrac{1}{12}\cos^{3}3x\cdot \mathrm{sen\,}3x+\dfrac{3}{4}\int{\cos^{2}3x\,dx}$


Fazendo $I_{2}=\int{\cos^{2}3x\,dx},$ temos

$I_{1}=\dfrac{1}{12}\cos^{3}3x\cdot \mathrm{sen\,}3x+\dfrac{3}{4}I_{2}$


$\bullet\;\;k=2:$

Rearranjando $I_{2}$:

$I_{2}=\int{\cos 3x\cdot \cos 3x\,dx}\\ \\ \\ \begin{array}{ll} u_{2}=\cos 3x\;\;&\;\;du_{2}=-3\mathrm{\,sen\,}3x\,dx\\ \\ dv_{2}=\cos 3x\,dx\;\;&\;\;v_{2}=\dfrac{1}{3}\mathrm{\,sen\,}3x \end{array}\\ \\ \\ \\ \int{u_{2}\,dv_{2}}=u_{2}v_{2}-\int{v_{2}\,du_{2}}\\ \\ \\ I_{2}=\dfrac{1}{3}\cos 3x\cdot \mathrm{sen\,}3x+\int{\mathrm{sen^{2}\,}3x\,dx}\\ \\ \\ I_{2}=\dfrac{1}{3}\cos 3x\cdot \mathrm{sen\,}3x+\int{(1-\cos^{2} 3x)\,dx}\\ \\ \\ I_{2}=\dfrac{1}{3}\cos 3x\cdot \mathrm{sen\,}3x+\int{dx}-\int{\cos^{2}3x\,dx}\\ \\ \\ I_{2}=\dfrac{1}{3}\cos 3x\cdot \mathrm{sen\,}3x+x-I_{2}\\ \\ \\ I_{2}+I_{2}=\dfrac{1}{3}\cos 3x\cdot \mathrm{sen\,}3x+x\\ \\ \\ 2I_{2}=\dfrac{1}{3}\cos 3x\cdot \mathrm{sen\,}3x+x$


$I_{2}=\dfrac{1}{6}\cos 3x\cdot \mathrm{sen\,}3x+\dfrac{1}{2}\,x$


Substituindo de volta em $I_{1}$, temos

$I_{1}=\dfrac{1}{12}\cos^{3}3x\cdot \mathrm{sen\,}3x+\dfrac{3}{4}\left(\dfrac{1}{6}\cos 3x\cdot \mathrm{sen\,}3x+\dfrac{1}{2}\,x \right )\\ \\ \\ I_{1}=\dfrac{1}{12}\cos^{3}3x\cdot \mathrm{sen\,}3x+\dfrac{1}{8}\cos 3x\cdot \mathrm{sen\,}3x+\dfrac{3}{8}\,x$


Substituindo de volta em $I_{0}$, temos

$I_{0}=\dfrac{1}{18}\cos^{5}3x\cdot \mathrm{sen\,}3x+\dfrac{5}{6}\left(\dfrac{1}{12}\cos^{3}3x\cdot \mathrm{sen\,}3x+\dfrac{1}{8}\cos 3x\cdot \mathrm{sen\,}3x+\dfrac{3}{8}\,x \right )\\ \\ \\ I_{0}=\dfrac{1}{18}\cos^{5}3x\cdot \mathrm{sen\,}3x+\dfrac{5}{72}\cos^{3}3x\cdot \mathrm{sen\,}3x+\dfrac{5}{48}\cos 3x\cdot \mathrm{sen\,}3x+\dfrac{5}{16}\,x+C$

que é o resultado procurado para a integral original.