Integral de e^(-ax)*cosbxdx dx
Solución
Respuesta (Indefinida)
[src]
// x for And(a = 0, b = 0)\
|| |
|| I*b*x I*b*x I*b*x |
|| x*cos(b*x)*e e *sin(b*x) I*x*e *sin(b*x) |
|| ----------------- + --------------- - ------------------- for a = -I*b |
/ || 2 2*b 2 |
| || |
| -a*x || -I*b*x -I*b*x -I*b*x |
| E *cos(b*x) dx = C + |
$$\int e^{- a x} \cos{\left(b x \right)}\, dx = C + \begin{cases} x & \text{for}\: a = 0 \wedge b = 0 \\- \frac{i x e^{i b x} \sin{\left(b x \right)}}{2} + \frac{x e^{i b x} \cos{\left(b x \right)}}{2} + \frac{e^{i b x} \sin{\left(b x \right)}}{2 b} & \text{for}\: a = - i b \\\frac{i x e^{- i b x} \sin{\left(b x \right)}}{2} + \frac{x e^{- i b x} \cos{\left(b x \right)}}{2} + \frac{e^{- i b x} \sin{\left(b x \right)}}{2 b} & \text{for}\: a = i b \\- \frac{a \cos{\left(b x \right)}}{a^{2} e^{a x} + b^{2} e^{a x}} + \frac{b \sin{\left(b x \right)}}{a^{2} e^{a x} + b^{2} e^{a x}} & \text{otherwise} \end{cases}$$
/ 1
| ---------- for And(2*|arg(b)| = 0, 2*|arg(a)| < pi)
| / 2\
| | b |
| a*|1 + --|
| | 2|
| \ a /
|
< oo
| /
| |
| | -a*x
| | cos(b*x)*e dx otherwise
| |
|/
|0
\
$$\begin{cases} \frac{1}{a \left(1 + \frac{b^{2}}{a^{2}}\right)} & \text{for}\: 2 \left|{\arg{\left(b \right)}}\right| = 0 \wedge 2 \left|{\arg{\left(a \right)}}\right| < \pi \\\int\limits_{0}^{\infty} e^{- a x} \cos{\left(b x \right)}\, dx & \text{otherwise} \end{cases}$$
=
/ 1
| ---------- for And(2*|arg(b)| = 0, 2*|arg(a)| < pi)
| / 2\
| | b |
| a*|1 + --|
| | 2|
| \ a /
|
< oo
| /
| |
| | -a*x
| | cos(b*x)*e dx otherwise
| |
|/
|0
\
$$\begin{cases} \frac{1}{a \left(1 + \frac{b^{2}}{a^{2}}\right)} & \text{for}\: 2 \left|{\arg{\left(b \right)}}\right| = 0 \wedge 2 \left|{\arg{\left(a \right)}}\right| < \pi \\\int\limits_{0}^{\infty} e^{- a x} \cos{\left(b x \right)}\, dx & \text{otherwise} \end{cases}$$
Piecewise((1/(a*(1 + b^2/a^2)), (2*Abs(arg(b)) = 0))∧(2*Abs(arg(a)) < pi), (Integral(cos(b*x)*exp(-a*x), (x, 0, oo)), True))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.