Question

Olá, como eu poderia resolver esta integral de potencias trigonometricas ?

Answer 1

$\int sen^3(x)*tg^{-5}(x) \; dx \\\\\boxed{tg(x)= \frac{sen(x)}{cos(x)} }\\\\\\ = \int sen^3(x)* \frac{sen^{-5}(x)}{cos^{-5}(x)} \;dx\\\\ =\int \frac{sen^{-2}(x)}{cos^{-5}(x)}\;dx = \int \frac{cos^5(x)}{sen^2(x)} \;dx\\\\\\\boxed{cos^2(x)+sen^2(x)=1}\\\\\\ \int \frac{(cos^2(x))^2*cos(x)}{sen^2(x)} \;dx = \int \frac{(1-sen^2(x))^2*cos(x)}{sen^2(x)} \;dx\\\\ = \int \frac{(1-2sen^2(x)+sen^4(x))*cos(x)}{sen^2(x)}\;dx \\\\ = \int \left ( \frac{1}{sen^2(x)}-2+sen^2(x)\right)*cos(x) \; dx$

substituição 
$u=sen(x)\\ du= cos(x)\;dx$


$\int ( \frac{1}{u^2}-2+u^2) du = \frac{-1}{u}-2u+ \frac{u^3}{3}+C\\\\ = \boxed{\boxed{ \frac{-1}{sen(x)}-2sen(x)+ \frac{sen^3(x)}{3} +C }}$