Question

integrate by parts
sin^(-1)xdx

Answer 1

try\[u=\sin^{-1}x\\dv=dx\]

Answer 2

I tried that and I always get stuck once I plug everything into the equation uv-integral vdu

Answer 3

what do you get when you plug in?

Answer 4

\[\sin ^{-1}x^2-\int\limits((x)/(1-x^2))dx\]

Answer 5

the integral is right, but how do you get sin^{-1}(x^2)

Answer 6

sin^(-1)x*x that was my u*v.

Answer 7

is it just xsin^(-1)x

Answer 8

right, not for that integral, do a *simple*, very common u-sub
u=(what's under the radical)

Answer 9

now for*

Answer 10

oh you forgot the radical

Answer 11

\[u=\sin^{-1}x\implies du=?\]

Answer 12

\[1/(\sqrt{1-x^2})\]

Answer 13

so \[\large\int vdu=?\]

Answer 14

\[(1/\sqrt{(1-x^2}))xdx\]

Answer 15

this is where i get stuck

Answer 16

good, so now it's a simple u-sub
let u=(what's under the radical)
then du=?

Answer 17

\[(1/2)(1-x^2)^{-1/2}*(-2x)dx\] ??

Answer 18

yes, now you can integrate it

Answer 19

still stuck?

Answer 20

it doesn't make sense to have du be that long of a derivative

Answer 21

why not?
simply substitute in what we used as our u-substitution in the second integral
u=1-x^2
du=-2xdx

Answer 22

what does that give when you substitute it into the integral?

Answer 23

\[xsin ^{-1}x-\int\limits(du/\sqrt{u)}\]
?

Answer 24

you dropped the -1/2

Answer 25

oh wait its \[xsin ^{-1}x-\int\limits(-1/2)du/(\sqrt{u)}dx\]

Answer 26

yes, except now you pulled a dx out from somewhere.... not sure why :P

Answer 27

on my paper i don't have the dx, sorry.

Answer 28

no worries, so we can pull out that constant, giving\[x\sin^{-1}x+\frac12\int {du\over \sqrt u}\]

Answer 29

which you should be able to integrate

Answer 30

I have that

Answer 31

ok, so what's the problem?

Answer 32

I know what du/u is, ln(u)+c, but not du/sqrt. u.

Answer 33

try writing 1/sqrtx as a fractional exponent