| x+sinx | ||
Proszę Państwa o jak najszybsza pomoc mam problem w calce ∫ | dx tam w | |
| sinx2 |
| 1 | ||
u = x+ sinx, v' = | ||
| sin2x |
| cosx(1+cosx) | ||
= −(x+ sinx)*ctg(x) + ∫ | = | |
| sinx |
| cosx+cos2x) | ||
= −(x+ sinx)*ctg(x) + ∫ | = | |
| sinx |
| cosxdx | cos2x) | |||
= −(x+ sinx)*ctg(x) + ∫ | +∫ | = | ||
| sinx | sinx |
| cosxdx | 1−sin2x | |||
= −(x+ sinx)*ctg(x) + ∫ | +∫ | = | ||
| sinx | sinx |
| cosxdx | dx | sin2x dx | ||||
= −(x+ sinx)*ctg(x) + ∫ | +∫ | − ∫ | = | |||
| sinx | sinx | sinx |
| cosxdx | dx | |||
= −(x+ sinx)*ctg(x) + ∫ | +∫ | − ∫sinxdx = | ||
| sinx | sinx |
| cosxdx | f(x)dx | |||
∫ | = korzystając ze wzoru: ∫ | = ln|f(x)| + C | ||
| sinx | f'(x) |
| 1 | sinx | sinx | ||||
∫ | dx = ∫ | dx = ∫ | dx = | t = cosx, dt = − sinxdx | = | |||
| sinx | sin2x | 1 − cos2x |
| f'(x) | ||
korzystając ze wzoru: ∫ | dx = ln|f(x)| + C | |
| f(x) |
| x+sinx | x | dx | ||||
I= ∫ | dx= ∫ | dx+∫ | dx= I1+I2(*) , gdzie | |||
| sin2x | sin2x | sinx |
| x | dx | dx | ||||
I1= ∫ | dx= |u=x, dv= | i du=dx i v= ∫ | = tgx|= | |||
| sin2x | sin2x | sin2x |
| −sinx | ||
= xtgx−∫tgxdx= xtg+∫ | dx= xtgx+ln|cosx| ; | |
| cosx |
| dx | ||
I2= ∫ | dx= ...= ln|tg12x| , zatem w (*) mamy | |
| sinx |
| x | sinx | |||
a to nie powinno się rozbic na całke ∫ | dx +∫ | dx | ||
| sin2x | sin2 |
| x | sinx | |||
a to nie powinno się rozbic na całke ∫ | dx +∫ | dx | ||
| sin2x | sin2 |