∫11−sin4xdx
∫ln(x+√1+x2)dx
| 1 | 1 | |||
...∫ | dx = ∫ | dx = ... | ||
| (1 − sin2x)(1 + sin2x) | cos2x(1 + sin2x) |
| dx | tg2x | |||
podstawienie: t = tgx , dt = | , sin2x = | |||
| cos2x | tg2x + 1 |
| t2 | ||
sin2x = | .... oczywiście... | |
| t2 + 1 |
| 1 | 1 | 1 | 1 | ||||
= | + | + | |||||
| 1−sin4x | 4(1−sinx) | 4(1+sinx) | 2(1+sin2x) |
| x | ||
zamień na tg | a potem ułamki proste | |
| 2 |
Fakt Funkcja nie zmienia znaku 
| dt | ||
tgx = t => x = arctg t, dx = | ||
| 1 + t2 |
| tg2x | ||
sin2x = | ||
| 1 + tg2x |