Question

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

Answer 1

$\displaystyle \int \tan^33x \,dx$

cambio de variable
$u=3x \to du = 3dx\to dx =\dfrac{1}{3}du$

$\displaystyle \int \tan^33x \,dx=\frac{1}{3}\int \tan^3 u \,du$

Sustitución trigonométrica.
Sea 
$z=\tan u\to u=\arctan z\to du=\dfrac{1}{1+z^2}\,dz$

$\displaystyle \int \tan^3 u \,du=\int z^3 \cdot \dfrac{1}{1+z^2}\,dz\\ \\ \int \tan^3 u \,du=\int z -\dfrac{z}{1+z^2}dz\\ \\ \int \tan^3 u \,du= \dfrac{z^2}{2}-\dfrac{1}{2}\ln|1+z^2|\\ \\ \int \tan^3 u \,du=\dfrac{\tan^2u}{2}-\dfrac{1}{2}\ln|1+\tan^2u|\\ \\ \int \tan^3 u \,du=\dfrac{\tan^2u}{2}-\dfrac{1}{2}\ln|\sec^2u|\\ \\ \int \tan^3 u \,du=\dfrac{\tan^2u}{2}-\ln|\sec x|\\ \\ \int \tan^3 u \,du=\dfrac{\tan^2u}{2}+\ln|\cos x|$

Entonces

$\displaystyle \int \tan^33x \,dx=\dfrac{\tan^23x}{6}+\dfrac{1}{3}\ln|\cos 3x|+C\\ \\ \int \tan^33x \,dx=\dfrac{\sec^23x-1}{6}+\dfrac{1}{3}\ln|\cos 3x|+C\\ \\ \int \tan^33x \,dx=\dfrac{\sec^23x}{6}+\dfrac{1}{3}\ln|\cos 3x|+C-\dfrac{1}{6}\\ \\ \\ \boxed{\int \tan^33x \,dx=\dfrac{\sec^23x}{6}+\dfrac{1}{3}\ln|\cos 3x|+K}$

Answer 2

∫tg³(3x)dx = ∫tg²(3x)tg(3x)dx = ∫(sec²3x -1)tg3xdx
∫sec²3xtg3xdx - ∫tg3xdx = ∫sec3xsec3xtg3xdx - ∫tg3xdx
__sec²3x__  +  __lncos3x_
      2×3                  3
Comprovação achando a derivada da primitiva acima:
_1_×2sec3xsec3xtg3xdx - __1__×_sen3x_(3)_dx___
  6                                     3      cos3x  
sec²3xtg3x - tg3x = tg3x(sec²3x - 1) = tg3x(tg²3x) = tg³3xdx