| dx | π | ||||||||||||
∫ | = − ctg(2x+ | ) + C czy to jest prawidłowe rozwiązanie | |||||||||||
| 4 |
| dx | ||
∫ | = | |
| 1+Y |
| x5 | ||
∫ | dx= | |
| √1−x2 |
| dx | ||
∫ | = −12*ctg(2x+π4) | |
| sin2(2x+π4) |
| dx | xdx | |||
dt = | *(−2x) = − | |||
| 2√1−x2 | √1−x2 |
| x5 | −x5 | x4*xdx | ||||
∫ | dx = −∫ | dx = −∫ | = | |||
| √1−x2 | √1−x2 | √1−x2 |
| xdx | ||
∫x4*(− | ) = ∫(1−2t2+t4) dt | |
| √1−x2 |
| 1 | ||
skąd w ad.1 ta | ||
| 2 |
| 1 | 2 | |||
[ctg(2x+π4)]' = − | *(2x+π4)' = − | |||
| sin2(2x+π4) | sin2(2x+π4) |
| 1 | |
= [−12ctg(2x+π4)]' | |
| sin2(2x+π4) |