Question

integral de sen^5 (2x)/cos^2 (2x) dx​

Answer 1

$\displaystyle \int \frac{\text{sen}^5(2\text x)\text {dx}}{\text{cos}^2(2\text x)} \\\\\\ \int \frac{[\text{sen}^2(2\text x)]^2.\text{sen}(2\text x)\text {dx}}{\text{cos}^2(2\text x)} \\\\\\ \int \frac{[\ 1-\text{cos}^2(2\text x)\ ]^2.\text{sen(2x)dx}}{\text{cos}^2(2\text x) } \\\\\\ \underline{\text{fa{\c c}amos}}: \\\\ \text u = 2\text x \to \text {dx}=\frac{\text {du}}{2} \\\\\\ \int \frac{[\ 1-\text{cos}^2(\text u)\ ]^2.\text{sen(u)du}}{2.\text{cos}^2(\text u) }$

$\displaystyle \frac{1}{2}.\int \frac{[\ 1-\text{cos}^2(\text u)\ ]^2.\text{sen(u)du}}{\text{cos}^2(\text u) } \\\\\\ \underline{\text {fa{\c c}amos}}: \\\\ \text v = \text{cos(u)} \to -\text{dv}=\text{sen(u).du} \\\\\\\frac{1}{2} \int \frac{-[\ 1-\text v^2\ ]^2\text {dv}}{\text v^2} \\\\\\ \frac{-1}{2} \int [\ \frac{1}{\text v^2} -\frac{2\text v^2}{\text v^2}+\frac{\text v^4}{\text v^2} \ ]\text{dv} \\\\\\ \frac{-1}{2}[\ \int \frac{1\text{dv}}{\text v^2} -\int 2\text{dv} +\int \text v^2\text{dv} \ ]$

$\displaystyle \frac{-1}{2} \ [ \ \frac{-1}{\text v}-2\text v + \frac{\text v^3}{3} \ ] + \text C \\\\\\ \frac{1}{2\text v} +\text v-\frac{\text v^3}{6}+ \text C$

Desfazendo a troca de variável :

$\displaystyle \frac{1}{2.\text{cos(2x)}}+\text{cos(2x)}-\frac{\text{cos}^3(2\text x)}{6} + \text C \\\\\\ \huge\boxed{\frac{\text{sec(2x)}}{2}+\text{cos(2x)}-\frac{\text{cos}^3(2\text x)}{6} + \text C }$