| cos2x | 1−sin2x | |||
∫ | *cosx dx = ∫ | *cosx dx= | ||
| sin3x | sin3x |
| 1−t2 | 1 | 1 | ||||
∫ | dt= ∫ | dt − ∫ | dt= ∫t−3 dt − ln|t|= | |||
| t3 | t3 | t |
| 1 | 1 | |||
=− | −ln|sinx|+c=− | −ln|sinx|+c | ||
| 2t2 | 2sin2x |
| ctg2x | ||
w odpowiedziach jest natomiast: − | −ln|sinx|+c | |
| 2 |
| ctg2x | cos2x | 1−sin2x | |||
+ c = | + c = | + c = | |||
| 2 | 2sin2x | 2sin2x |
| 1 | 1 | 1 | ||||
= | − | + c = | + c1 | |||
| 2sin2x | 2 | 2sin2x |
