| dx | dt | |||
∫ | , niech tgx = t → arctgt = x → | = dx, korzystając z | ||
| sin2x + tg2x | t2+1 |
| tg2x | t2 | |||
tożsamości odkrywasz że sin2x = | → sin2x = | |||
| 1+tg2x | 1+ t2 |
| dx | dt | |||
∫ | → ∫ | = | ||
| sin2x + tg2x | t4 + 2t2 |
| dt | 1 | dt | dt | |||||
= ∫ | = | (∫ | − ∫ | ) | ||||
| t2(t2 + 2) | 2 | t2 | t2+2 |
| 1 | 1 | 1 | t | |||||
= | (− | − | arctg( | ) | ||||
| 2 | t | √2 | √2 |
| dx | 1 | 1 | 1 | tgx | ||||||
∫ | = | (− | − | arctg( | ) | |||||
| sin2x + tg2x | 2 | tgx | √2 | √2 |
| dx | 1 | 1 | tgx | |||||
∫ | = | (−ctgx − | arctg( | ) | ||||
| sin2x + tg2x | 2 | √2 | √2 |
| dx | ||
∫ | = | |
| (ex + e−x)2 |
| tdt | ||
∫ | = [podstawienie t2+1 = k →2tdt = dk] = | |
| (t2+1)2 |
| 1 | dk | 1 | 1 | 1 | |||||
= | ∫ | = − | k−1 = − | → | |||||
| 2 | k2 | 2 | 2 | t2+1 |
| dx | 1 | 1 | |||
∫ | = − | ||||
| (ex + e−x)2 | 2 | (ex)2+1 |