Question

Como resolvo esta integral?

Answer 1

$\boxed{ \int \frac{sec^6(\theta)}{tg^2(\theta)} d\theta}$

aplicando a substituiçao
$u=tg(\theta)\\\\du =sec^2(\theta).d\theta$

$\int \frac{sec^4(\theta)}{u^2} du$

lembrando que 
$\boxed{sec^2(\theta) = tg^2(\theta)+1}$

então 
$sec^4(\theta) = (tg^2(\theta)+1)^2 = (u^2+1)^2$

ficando com 
$\int { \frac{(u^2+1)^2}{u^2} } \, du \\\\ = \int \frac{u^4+2u^2+1}{u^2} du\\\\= \int \frac{u^4}{u^2} + \frac{2u^2}{u^2} + \frac{1}{u^2} du\\\\\\\int u^2 +2+ \frac{1}{u^2}du= \frac{u^3}{3}+2u - \frac{1}{u}+C$

substituindo o valor de u
$\frac{tg^3(\theta)}{3}+ 2*tg(\theta)- \frac{1}{tg(\theta)}+C$

logo
$\boxed{\int \frac{sec^6(\theta)}{tg^2(\theta)} d\theta =\frac{tg^3(\theta)}{3}+ 2*tg(\theta)- cotg(\theta)+C}$

Answer 2

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{sec^6(\theta)}{tg^2(\theta)}\,d\theta=\int\dfrac{(sec^2(\theta))^2\cdot sec^2(\theta)}{tg^2(\theta)}\,d\theta\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf u=tg(\theta)\longrightarrow du=sec^2(\theta)\,d\theta\\\sf vale~lembrar~que~sec^2(\theta)=1+tg^2(\theta)\\\displaystyle\sf\int\dfrac{(sec^2(\theta))^2\cdot sec^2(\theta)}{tg^2(\theta)}\,d\theta=\int\dfrac{(1+u^2)^2}{u^2}du\\\displaystyle\sf\int u^{-2}(1+2u^2+u^4)~du=\int( u^{-2}+2+u^2)du\end{array}}$

$\boxed{\begin{array}{l}\sf=-\dfrac{1}{u}+2u+\dfrac{1}{3}u^3+c\\\sf substituindo~temos\\\\\sf -\dfrac{1}{tg(\theta)}+2tg(\theta)+\dfrac{1}{3}tg^3(\theta)+k\end{array}}$