| sin3xdx | t = cosx | |||
∫ | = | | = | ||
| cos2x+2cosx + 5 | dt = −sinxdx |
| (−1)sin2x*(−sinxdx) | ||
∫ | = | |
| cos2x + 2cosx + 5 |
| (−1)*(1−cos2x)*(−sinxdx) | ||
∫ | = | |
| cos2x + 2cosx + 5 |
| (cos2x−1)*(−sinxdx) | ||
∫ | = podstawienie | |
| cos2x + 2cosx + 5 |
| (t2−1)*dt) | ||
∫ | = dzielenie: | |
| t2 + 2t + 5 |
| 2t+6 | 2t+2+4 | |||
∫[1 − | dt] = ∫dt − ∫ | dt = | ||
| t2+2t+5 | t2+2t+5 |
| dt | ||
t − ln(t2+2t+5) − 4*∫ | = | |
| t2+2t+5 |
| dt | ||
t − ln(t2+2t+5) − 4*∫ | = | |
| t2+2t+1 + 4 |
| dt | ||
t − ln(t2+2t+5) − 4*∫ | = | |
| (t+1)2 + 22 |
| t + 1 | ||
t − ln(t2+2t+5) − 2arctg( | ) + C | |
| 2 |
(sobie gdzies to zapisze)
4cosx + cos2x + 11 = 4cosx + cos2−sin2x + 11 = cos2x − (1−sin2x) + 4cosx + 11 =
2cos2x + 4cosx + 10 = 2(cos2x + 2cosx + 5)
(w liczniku jeszcze jest 2* ...)