Integral de 1/(sin(2*x))^3 dx
Solución
Respuesta (Indefinida)
[src]
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| 1 log(1 + cos(2*x)) log(-1 + cos(2*x)) cos(2*x)
| --------- dx = C - ----------------- + ------------------ + --------------------
| 3 8 8 / 2 \
| sin (2*x) 2*\-2 + 2*cos (2*x)/
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$$\int \frac{1}{\sin^{3}{\left(2 x \right)}}\, dx = C + \frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{8} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{8} + \frac{\cos{\left(2 x \right)}}{2 \left(2 \cos^{2}{\left(2 x \right)} - 2\right)}$$
log(1 - cos(4)) log(1 + cos(6)) log(1 - cos(6)) log(1 + cos(4)) cos(6) cos(4)
- --------------- - --------------- + --------------- + --------------- + ------------------ - ------------------
8 8 8 8 / 2 \ / 2 \
2*\-2 + 2*cos (6)/ 2*\-2 + 2*cos (4)/
$$\frac{\cos{\left(6 \right)}}{2 \left(-2 + 2 \cos^{2}{\left(6 \right)}\right)} + \frac{\log{\left(1 - \cos{\left(6 \right)} \right)}}{8} - \frac{\cos{\left(4 \right)}}{2 \left(-2 + 2 \cos^{2}{\left(4 \right)}\right)} + \frac{\log{\left(\cos{\left(4 \right)} + 1 \right)}}{8} - \frac{\log{\left(\cos{\left(6 \right)} + 1 \right)}}{8} - \frac{\log{\left(1 - \cos{\left(4 \right)} \right)}}{8}$$
=
log(1 - cos(4)) log(1 + cos(6)) log(1 - cos(6)) log(1 + cos(4)) cos(6) cos(4)
- --------------- - --------------- + --------------- + --------------- + ------------------ - ------------------
8 8 8 8 / 2 \ / 2 \
2*\-2 + 2*cos (6)/ 2*\-2 + 2*cos (4)/
$$\frac{\cos{\left(6 \right)}}{2 \left(-2 + 2 \cos^{2}{\left(6 \right)}\right)} + \frac{\log{\left(1 - \cos{\left(6 \right)} \right)}}{8} - \frac{\cos{\left(4 \right)}}{2 \left(-2 + 2 \cos^{2}{\left(4 \right)}\right)} + \frac{\log{\left(\cos{\left(4 \right)} + 1 \right)}}{8} - \frac{\log{\left(\cos{\left(6 \right)} + 1 \right)}}{8} - \frac{\log{\left(1 - \cos{\left(4 \right)} \right)}}{8}$$
-log(1 - cos(4))/8 - log(1 + cos(6))/8 + log(1 - cos(6))/8 + log(1 + cos(4))/8 + cos(6)/(2*(-2 + 2*cos(6)^2)) - cos(4)/(2*(-2 + 2*cos(4)^2))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.