| dx | ||
∫ | = | |
| 1+ctgx |
| cosx | 1−t2 | |||
przez podstawienie: ctgx= | = | i t=tgx2, | ||
| sinx | 2t |
| 2dt | ||
czyli x2=arctgt ⇒ x=2arctgt ⇒ dx= | , oraz | |
| 1+t2 |
| 1−t2 | 1+2t−t2 | |||
1+ctgx = 1+ | = | , wtedy dana całka | ||
| 2t | 2t |
| 2*2tdt | 4tdt | |||
= ∫ | = ∫ | = dalej całka wymierna | ||
| (1+2t−t2)(1+t2) | (1+2t−t2)(1+t2) |
| 4t | Ax+B | Cx+D | ||||
gdzie | = | + | = baw się sam(a).  | |||
| (1+2t−t2)(1+t2) | 1+2t−t2 | 1+t2 |