Question

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

Answer 1

   $u=\cot 2x \to x= \frac{1}{2}\mbox{arccot } u\to =dx =-\dfrac{1/2}{1+u^2}du\\ \\ \displaystyle \int \cot ^4 2x \,dx = -\dfrac{1}{2}\int u^4\cdot \dfrac{1}{1+u^2}du\\ \\ \int \cot ^4 2x \,dx = -\dfrac{1}{2}\int \dfrac{u^4}{1+u^2}du\\ \\ \int \cot ^4 2x \,dx = -\dfrac{1}{2}\int u^2-1+\dfrac{1}{1+u^2}du\\ \\ \int \cot ^4 2x \,dx = -\dfrac{1}{6}u^3+\dfrac{1}{2}u+\dfrac{1}{2}\mbox{arccot } u+C\\ \\ \boxed{\int \cot ^4 2x \,dx = -\dfrac{1}{6}\cot^32x+\dfrac{1}{2}\cot 2x+x+C}$

Answer 2

∫cotg^4(2x)dx = ∫cotg²2x(cosec²2x - 1)dx
∫cotg²(2x)cosec²(2x)dx - ∫cotg²2xdx
∫cotg²(2x)cosec²(2x)dx - ∫[(cosec²(2x) - 1])dx
∫cotg²(2x)cosec²(2x)dx  -∫cosec²(2x)dx  - ∫dx
- _cotg³(2x)_ +  _cotg(2x)_ + x  + C
     3(2)                 2
- _cotg³(2x)_ +  _cotg(2x)_ + x  + C
        6                   2