Question

( Cálculo I ): Resolva a integral:

$\large\displaystyle\text{ \begin{aligned} \int {\frac{\tan^3 (3x)}{\sqrt{\sec (3x)} } } \, dx \end{aligned} }$

Answer 1

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{tg^3(3x)}{\sqrt{sec(3x)}}~dx=\dfrac{1}{3}\int\dfrac{tg^2(3x)\cdot tg(3x)3}{\sqrt{sec(3x)}}~dx\\\underline{\rm multiplica~e~divide~por~sec(3x):}\\\displaystyle\sf\dfrac{1}{3}\int\dfrac{tg^2(3x)\cdot sec(3x)\cdot tg(3x)\cdot 3}{sec(3x)\cdot\sqrt{sec(3x)}}dx\\\underline{\rm usando~a~relac_{\!\!,}\tilde ao~tg^2(3x)=sec^2(3x)-1:}\\\displaystyle\sf\dfrac{1}{3}\int\dfrac{(sec^2(3x)-1)\cdot sec(3x)\cdot tg(3x)\cdot 3}{sec(3x)\sqrt{sec(3x)}}dx\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\boldsymbol{fac_{\!\!,}a}}\\\rm u= sec(3x)\implies du=sec(3x)\cdot tg(3x)\cdot 3dx\\\displaystyle\sf\dfrac{1}{3}\int\dfrac{(u^2-1)}{u\sqrt{u}}du=\dfrac{1}{3}\int(u^2-1)u^{-\frac{3}{2}}du\\\displaystyle\sf\dfrac{1}{3}\int(u^{\frac{1}{2}}-u^{-\frac{3}{2}})du=\dfrac{1}{3}\cdot\dfrac{2}{3}u^{\frac{3}{2}}-\dfrac{1}{3}\cdot-2 u^{-\frac{1}{2}}+k\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf\dfrac{1}{3}\int( u^{\frac{1}{2}}-u^{-\frac{3}{2}})du=\dfrac{2}{9} \sqrt[\sf3]{\sf u^2}+\dfrac{2}{3\sqrt{u}}+k\\\underline{\rm substituindo~podemos~concluir~que}\\\displaystyle\sf\int\dfrac{tg^3(3x)}{\sqrt{sec(3x)}}dx=\dfrac{2\sqrt[\sf3]{\sf sec^2(3x)}}{9}+\dfrac{2}{3\sqrt{sec(3x)}}+k\end{array}}$