Question

Cálculo integral simples

Answer 1

$\frac{1}{64} \int\limits {1} \, dr + \frac{1}{64} \int\limits {cos(r)} \, dr$Resposta:

$\int\limits^} {sen(2x)^{2}.cos(2x)^4} \, dx \\\int\limits^}(1-cos(2x)^2).cos(2x)^4dx\\\\\int\limits^}cos(2x)^4 -cos(2x)^6dx\\\int\limits^}cos(2x)^4dx -\int\limits^}cos(2x)^6dx$

$u = 2x \\\frac{du}{dx} = 2 \\\frac{du}{2} = dx$

$\frac{1}{2} \int\limits^}cos(u)^4du -\frac{1}{2} \int\limits {cos(u)^6} \, dx \\\\cos(u)^2 = \frac{1+cos(2u)}{2}$

$\frac{1}{8} \int\limits^}1du +\frac{1}{8} \int\limits^}cos(2u)^2du +\frac{1}{4} \int\limits^} cos(2u)du$

$y = 2u \\\frac{dy}{du} = 2 \\du= \frac{dy}{2}$

$\frac{1}{16}\int\limits^}1dy + \frac{1}{16}\int\limits^} cos(y)^2 dy +\frac{1}{4}\int\limits^}cos(y)dy$

$r = 2y\\dr=\frac{dy}{2}$

$\frac{1}{64} \int\limits {1} \, dr +\frac{1}{64} \int\limits cos(r)\, dr$
tentei meu chapa mais foi ate ais mesmo