całki
gosiata: ∫cos5x √sinxdx
23 sty 20:08
Dawid: ∫√sinx(1−sin2x)2cosxdx
podstawienie
23 sty 20:15
pigor: ..., niech
√sinx=t i t >0 ⇒
sinx=t2 i cosxdx=2tdt,
to ∫cos
5x
√sinxdx= ∫(1−sin
2x)
2*
√sinx*cosxdx= ∫(1−t
4)
2*t*2tdt=
= 2∫t
2(1−2t
4+t
8)dt= 2∫(t
2−2t
6+2t
8)dt=
= 2(
13t
3−
27t
7+
29t
9) +C=
2t3(13−27t4+29t6) +C=
=
2sinx√sinx (13−27sin2x+29sin3x) +C ...
23 sty 20:29