Question

in Substitution Rule

Answer 1

\[\int\limits\limits_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x}} dx\]

Answer 2

i would make the substitution:
x = sin^2(u)
dx = 2sin*cos du

\[\rightarrow 2\int\limits_{0}^{.5}u^{2} \sin(u) du\]
Then use integration by parts

Answer 3

but I don't know how to do it

Answer 4

Do you know integration by parts?

Answer 5

yeah i do but that question is on substitution not in integration by parts

Answer 6

Why can't we do both?

Answer 7

you can do
:)

Answer 8

it

Answer 9

\[\int\limits_{}^{}u^2 \sin(u) du=u^2(- \cos(u))-\int\limits_{}^{}2u (-\cos(u)) du\]
you will need to do integration by parts one more time

Answer 10

ok
Thank you a lot guys :)

Answer 11

one thing I am confused on here, where did we get u^2 ?
are you saying arcsin(sin^2u)=u^2 ?

Answer 12

yes
arcsin(sin u) = u by definition of the inverse function

Answer 13

somehow the idea that that extends to arcsin(sin^n(u))=u^n has never been brought to my attention.
thanks

Answer 14

could we have just used the substitution:\[u = \sin^{-1}(x)\]

Answer 15

ah disregard, i didnt look at the denominator properly.

Answer 16

yeah i don't know if that is true turing

Answer 17

yeah it can't be

Answer 18

arcsin(sin(u^n))=u^n
arcsin(sin^n(u))=?

Answer 19

i agree with the first equation you have turing

Answer 20

wait now that i look at it again, i think i have it backwards.
sorry everyone

sin^2(arcsin u) = u^2 , thats what i thought it was

Answer 21

I got

Answer 22

Oh man, I thought I was loosing it :/

Answer 23

i got