Integral de sec³xtan²x dx
Solución
Respuesta (Indefinida)
[src]
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| 3
| 3 2 log(1 + sin(x)) log(-1 + sin(x)) sin (x) + sin(x)
| sec (x)*tan (x) dx = C - --------------- + ---------------- + --------------------------
| 16 16 2 4
/ 8 - 16*sin (x) + 8*sin (x)
$$\int \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)}\, dx = C + \frac{\sin^{3}{\left(x \right)} + \sin{\left(x \right)}}{8 \sin^{4}{\left(x \right)} - 16 \sin^{2}{\left(x \right)} + 8} + \frac{\log{\left(\sin{\left(x \right)} - 1 \right)}}{16} - \frac{\log{\left(\sin{\left(x \right)} + 1 \right)}}{16}$$
3
log(1 + sin(1)) log(1 - sin(1)) sin (1) + sin(1)
- --------------- + --------------- + --------------------------
16 16 2 4
8 - 16*sin (1) + 8*sin (1)
$$\frac{\log{\left(1 - \sin{\left(1 \right)} \right)}}{16} - \frac{\log{\left(\sin{\left(1 \right)} + 1 \right)}}{16} + \frac{\sin^{3}{\left(1 \right)} + \sin{\left(1 \right)}}{- 16 \sin^{2}{\left(1 \right)} + 8 \sin^{4}{\left(1 \right)} + 8}$$
=
3
log(1 + sin(1)) log(1 - sin(1)) sin (1) + sin(1)
- --------------- + --------------- + --------------------------
16 16 2 4
8 - 16*sin (1) + 8*sin (1)
$$\frac{\log{\left(1 - \sin{\left(1 \right)} \right)}}{16} - \frac{\log{\left(\sin{\left(1 \right)} + 1 \right)}}{16} + \frac{\sin^{3}{\left(1 \right)} + \sin{\left(1 \right)}}{- 16 \sin^{2}{\left(1 \right)} + 8 \sin^{4}{\left(1 \right)} + 8}$$
-log(1 + sin(1))/16 + log(1 - sin(1))/16 + (sin(1)^3 + sin(1))/(8 - 16*sin(1)^2 + 8*sin(1)^4)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.