oblicz całkę metodą podstawiania
wojtas2212: ∫tg(−3x+5)dx=
6 lis 19:59
karty do gry : t = cos(−3x + 5)
6 lis 20:02
Mariusz:
Można też podstawić za całego tangensa jeśli wynik chcemy mieć wyrażony za pomocą tangensa
| | tg(x+Δx)−tg(x) | |
limΔx→0 |
| = |
| | Δx | |
| | | tg(x)+tg(Δx) | |
| −tg(x) | | 1−tg(x)tg(Δx) | |
| |
limΔx→0 |
| = |
| | Δx | |
| | | tg(x)+tg(Δx)−tg(x)(1−tg(x)tg(Δx)) | |
| | | 1−tg(x)tg(Δx) | |
| |
limΔx→0 |
| = |
| | Δx | |
| | | tg(x)+tg(Δx)−tg(x)+tg2(x)tg(Δx) | |
| | | 1−tg(x)tg(Δx) | |
| |
limΔx→0 |
| = |
| | Δx | |
| | | tg(Δx)+tg2(x)tg(Δx) | |
| | | 1−tg(x)tg(Δx) | |
| |
limΔx→0 |
| = |
| | Δx | |
| | | tg(Δx)(1+tg2(x)) | |
| | | 1−tg(x)tg(Δx) | |
| |
limΔx→0 |
| = |
| | Δx | |
| | tg(Δx)(1+tg2(x)) | 1 | |
limΔx→0 |
|
| = |
| | 1−tg(x)tg(Δx) | Δx | |
| | tg(Δx) | 1+tg2(x) | |
limΔx→0 |
|
| = |
| | Δx | 1−tg(x)tg(Δx) | |
| | sin(Δx) | 1 | 1+tg2(x) | |
limΔx→0 |
|
|
| = |
| | Δx | cos(Δx) | 1−tg(x)tg(Δx) | |
| | 1+tg2(x) | |
limΔx→0 |
| |
| | 1−tg(x)tg(Δx) | |
=1+tg
2(x)
6 lis 20:30