Integral de X*sinx*sinkx dx
Solución
Respuesta (Indefinida)
[src]
// 2 2 2 2 2 \
|| cos (x) x *cos (x) x *sin (x) x*cos(x)*sin(x) |
|| ------- - ---------- - ---------- + --------------- for k = -1|
|| 4 4 4 2 |
|| |
/ || 2 2 2 2 2 |
| || cos (x) x *cos (x) x *sin (x) x*cos(x)*sin(x) |
| x*sin(x)*sin(k*x) dx = C + |< - ------- + ---------- + ---------- - --------------- for k = 1 |
| || 4 4 4 2 |
/ || |
|| 2 2 3 |
||sin(x)*sin(k*x) k *sin(x)*sin(k*x) x*cos(x)*sin(k*x) 2*k*cos(x)*cos(k*x) k*x*cos(k*x)*sin(x) x*k *cos(x)*sin(k*x) x*k *cos(k*x)*sin(x) |
||--------------- + ------------------ - ----------------- + ------------------- + ------------------- + -------------------- - -------------------- otherwise |
|| 4 2 4 2 4 2 4 2 4 2 4 2 4 2 |
\\ 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k /
∫xsin(x)sin(kx)dx=C+⎩⎨⎧−4x2sin2(x)−4x2cos2(x)+2xsin(x)cos(x)+4cos2(x)4x2sin2(x)+4x2cos2(x)−2xsin(x)cos(x)−4cos2(x)−k4−2k2+1k3xsin(x)cos(kx)+k4−2k2+1k2xsin(kx)cos(x)+k4−2k2+1k2sin(x)sin(kx)+k4−2k2+1kxsin(x)cos(kx)+k4−2k2+12kcos(x)cos(kx)−k4−2k2+1xsin(kx)cos(x)+k4−2k2+1sin(x)sin(kx)fork=−1fork=1otherwise
/ 2
| 1 sin (1) cos(1)*sin(1)
| - - - ------- + ------------- for k = -1
| 4 4 2
|
| 2
| 1 sin (1) cos(1)*sin(1)
< - + ------- - ------------- for k = 1
| 4 4 2
|
| 2 2 3
| 2*k sin(1)*sin(k) cos(1)*sin(k) k*cos(k)*sin(1) k *cos(1)*sin(k) k *sin(1)*sin(k) k *cos(k)*sin(1) 2*k*cos(1)*cos(k)
|- ------------- + ------------- - ------------- + --------------- + ---------------- + ---------------- - ---------------- + ----------------- otherwise
| 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2
\ 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k
⎩⎨⎧−41−4sin2(1)+2sin(1)cos(1)−2sin(1)cos(1)+4sin2(1)+41−k4−2k2+1k3sin(1)cos(k)+k4−2k2+1k2sin(k)cos(1)+k4−2k2+1k2sin(1)sin(k)+k4−2k2+1ksin(1)cos(k)+k4−2k2+12kcos(1)cos(k)−k4−2k2+12k−k4−2k2+1sin(k)cos(1)+k4−2k2+1sin(1)sin(k)fork=−1fork=1otherwise
=
/ 2
| 1 sin (1) cos(1)*sin(1)
| - - - ------- + ------------- for k = -1
| 4 4 2
|
| 2
| 1 sin (1) cos(1)*sin(1)
< - + ------- - ------------- for k = 1
| 4 4 2
|
| 2 2 3
| 2*k sin(1)*sin(k) cos(1)*sin(k) k*cos(k)*sin(1) k *cos(1)*sin(k) k *sin(1)*sin(k) k *cos(k)*sin(1) 2*k*cos(1)*cos(k)
|- ------------- + ------------- - ------------- + --------------- + ---------------- + ---------------- - ---------------- + ----------------- otherwise
| 4 2 4 2 4 2 4 2 4 2 4 2 4 2 4 2
\ 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k 1 + k - 2*k
⎩⎨⎧−41−4sin2(1)+2sin(1)cos(1)−2sin(1)cos(1)+4sin2(1)+41−k4−2k2+1k3sin(1)cos(k)+k4−2k2+1k2sin(k)cos(1)+k4−2k2+1k2sin(1)sin(k)+k4−2k2+1ksin(1)cos(k)+k4−2k2+12kcos(1)cos(k)−k4−2k2+12k−k4−2k2+1sin(k)cos(1)+k4−2k2+1sin(1)sin(k)fork=−1fork=1otherwise
Piecewise((-1/4 - sin(1)^2/4 + cos(1)*sin(1)/2, k = -1), (1/4 + sin(1)^2/4 - cos(1)*sin(1)/2, k = 1), (-2*k/(1 + k^4 - 2*k^2) + sin(1)*sin(k)/(1 + k^4 - 2*k^2) - cos(1)*sin(k)/(1 + k^4 - 2*k^2) + k*cos(k)*sin(1)/(1 + k^4 - 2*k^2) + k^2*cos(1)*sin(k)/(1 + k^4 - 2*k^2) + k^2*sin(1)*sin(k)/(1 + k^4 - 2*k^2) - k^3*cos(k)*sin(1)/(1 + k^4 - 2*k^2) + 2*k*cos(1)*cos(k)/(1 + k^4 - 2*k^2), True))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.