Question

Integral de tangente elevado a quinta potencia de x multiplicado por cós elevado ao quadrado de x

Answer 1

Calcular a integral indefinida

     ∫ tg⁵ x · cos² x dx.


Aplicando a definição de tangente, podemos escrever a integral como

     $\displaystyle\int \mathrm{tg^5\,}x\cdot \cos^2 x\,dx\\\\\\ =\int\bigg(\frac{\mathrm{sen\,}x}{\cos x}\bigg)^{\!5}\cdot \cos^2 x\,dx\\\\\\ =\int\frac{\mathrm{sen^5\,}x}{\cos^5 x}\cdot \cos^2 x\,dx\\\\\\ =\int\frac{\mathrm{sen^5\,}x}{\cos^3 x\cdot \cos^2 x}\cdot \cos^2 x\,dx\\\\\\ =\int\frac{\mathrm{sen^5\,}x}{\cos^3 x}\,dx$

     $\displaystyle=\int\frac{\mathrm{sen^4\,}x}{\cos^3 x}\cdot \mathrm{sen\,}x\,dx\\\\\\ =\int\frac{(\mathrm{sen^2\,}x)^2}{\cos^3 x}\cdot \mathrm{sen\,}x\,dx\\\\\\ =\int\frac{(1-\cos^2 x)^2}{\cos^3 x}\cdot \mathrm{sen\,}x\,dx\\\\\\ =-\int\frac{(1-\cos^2 x)^2}{\cos^3 x}\cdot (-\,\mathrm{sen\,}x)\,dx$


Faça a seguinte substituição:

     $\cos x=u\quad\Longrightarrow\quad -\,\mathrm{sen\,}x\,dx=du$


e a integral fica

     $\displaystyle=-\int\frac{(1-u^2)^2}{u^3}\,du$


Expanda o quadrado da diferença no numerador:

     $\displaystyle=-\int\frac{1^2-2\cdot 1\cdot u^2+ (u^2)^2}{u^3}\,du\\\\\\ =-\int\frac{1-2u^2+u^4}{u^3}\,du$


Separe as frações e simplifique as potências:

     $\displaystyle=-\int\bigg(\frac{1}{u^3}-\frac{2u^2}{u^3}+\frac{u^4}{u^3}\bigg)du\\\\\\ =-\,\bigg(\int\frac{1}{u^3}\,du-2\int\frac{u^2}{u^3}\,du+\int\frac{u^4}{u^3}\,du\bigg)\\\\\\ =-\int\frac{1}{u^3}\,du+2\int\frac{u^2}{u^3}\,du-\int\frac{u^4}{u^3}\,du\\\\\\ =-\int u^{-3}\,du+2\int u^{2-3}\,du-\int u^{4-3}\,du\\\\\\ =-\int u^{-3}\,du+2\int u^{-1}\,du-\int u^1\,du$


Agora é só aplicar a regra para achar primitivas de potências:

     $\displaystyle\int u^n\,du=\left\{\! \begin{array}{ll} \dfrac{u^{n+1}}{n+1}+C&\quad\textsf{se~~}n\ne -1\\\\ \ln|u|+C&\quad\textsf{se~~}n=-1 \end{array} \right.$


e a integral fica

     $=-\,\dfrac{u^{-3+1}}{-3+1}+2\ln|u|-\dfrac{u^{1+1}}{1+1}+C\\\\\\ =-\,\dfrac{u^{-2}}{-2}+2\ln|u|-\dfrac{u^2}{2}+C\\\\\\ =\dfrac{1}{2u^2}+2\ln|u|-\dfrac{u^2}{2}+C$


Substitua de volta para a variável  x:

     $=\dfrac{1}{2\cos^2 x}+2\ln|\!\cos x|-\dfrac{\cos^2 x}{2}+C\quad\longleftarrow\quad\mathsf{esta~\acute{e}~a~resposta.}$


Bons estudos! :-)