Question

Help with integral

Answer 1

int sin^5(x) / sqrt(1-x^2)

Answer 2

@zepdrix

Answer 3

@satellite73

Answer 4

try
x=cos(t)

Answer 5

-sin(cos(t))dt after substitution

Answer 6

no, you changed the exponent

Answer 7

in the denominator you get sin(t), and dx=-sin(t)dt

Answer 8

-sin^5(cost) dt

Answer 9

right, but we will now need another substitution, so I suggest you keep it in the form
sin^4(cost)(-sin(t))dt
now you can use a u-substitution

Answer 10

shouldn't it be sin^4(cost)*(sin(cost))dt ?

Answer 11

oh hm... you are right, we messed up earlier
after the sub it becomes
int sin^5(cos(t))/sqrt(1-cos^2(t)) *(-sin(t))dt
=int -sin^5(cos(t))

hm... darn. gotta keep thinking I guess

Answer 12

question 1: where did this integral come from?
question 2: are you sure you want the indefinite integral?

Answer 13

from exam. teacher said i should use some smart substitution

Answer 14

yes the indefinite one

Answer 15

but even wolfram can't do this, so maybe my teacher made a mistake

Answer 16

i tried whole bunch of substitution, neither one works

Answer 17

yeah, I've noticed that. No matter how I slice it, wolfram won't do it. Doesn't mean it can't be done, but it's discouraging

Answer 18

oh well, I'll just keep meditating on that magic substitution

Answer 19

ok, thanks

Answer 20

you tell me if you see a mistake in this one:

u=sqrt(1-x^2) -> x=sqrt(1-u^2)
dx=-2udu/sqrt(1-u^2)

integral becomes

int sin^5[sqrt(1-u^2)](-2u)du/[u*sqrt(1-u^2)]
=-int sin^5[sqrt(1-u^2)]udu/sqrt(1-u^2)

v=sqrt(1-u^2) ->dv=-2udu/sqrt(1-u^2) -> -dv/2=udu/sqrt(1-u^2)

integral becomes

1/2 int sin^5(v)dv

did I make a mistake?

Answer 21

int sin^5[sqrt(1-u^2)](-2u)du/[u*sqrt(1-u^2)]
=-int sin^5[sqrt(1-u^2)]du/sqrt(1-u^2)

u's cancel out

Answer 22

and dx=-udu/sqrt(1-u^2)

Answer 23

but that's a detail

Answer 24

fudge, yep this one is gnarly