Question

how to integrate

\(\int (\sin^{-1} x)^{2} dx\)

Answer 1

An approach to this is with integral by parts

Answer 2

u = arcsin(x)
du = 1/(1-x^(2))dx
du(1-x^(2)) = dx

so now you have

(u)^(2)(1-x^(2))

Answer 3

sorry

\[\int\limits_{}^{}(u)^{2}\sqrt{1-x^{2}}\]

Answer 4

it should be
u = arcsin(x)
du = 1/(1-x^(2))^(1/2)dx
du(1-x^(2))^(1/2) = dx

Answer 5

Now solve for x and use integration by parts

Answer 6

Solve for x
u = arcsin(x)
sin(u) = sin(arcsin(x))
sin(u) = x

Answer 7

Now we have

\[\int\limits_{}^{}u^{2}\sqrt{1-\sin^{2}(u)}\]

use the identity 1 - sin^(2)(u) = cos^(2)

now we have

\[\int\limits\limits_{}^{}u^{2}\cos(u)\]

Answer 8

Use substitution by parts

Answer 9

no, you were right..

Answer 10

oh yeah :)

Answer 11

Thanks

Answer 12

keep going, I am watching you (:

Answer 13

Now use substitution by parts
\[\int\limits\limits\limits_{}^{}u^{2}\cos(u)du = u^{2}\sin(u) - \int\limits_{}^{}2usin(u)dx\]

Answer 14

you mean du

Answer 15

yeah ignore my sloppy notation :)

Answer 16

use substitution by parts again

\[u^{2}\sin(u) + 2ucos(u) + 2\int\limits\limits\limits_{}^{}\cos(u)du\]

so

u^(2)sin(u) + 2ucos(u) - 2sin(u) + c

Now we need to substitute

Answer 17

u^(2)sin(u)- 2ucos(u) - 2sin(u) + c ?

Answer 18

arcsin^(2)(x)x + 2arcsin(x)cos(arcsin(x)) - 2x + c

Answer 19

i think there's a wrong sign there :P

Answer 20

the anti-derivative of sin(x) is -cos(x)

Answer 21

the antiderivative of cosx is sinx though haha

Answer 22

No I'm pretty sure I'm right it seems right to me

Answer 23

\[u^{2}\sin(u) - 2u(-\cos(u)) - 2\int\limits_{}^{} -\cos(u)du\]

Answer 24

I really enjoyed that thanks :)

Answer 25

haha lol =))

Answer 26

arcsin^(2)(x)x + 2(1-x^(2))^(1/2)arcsin(x) - 2x + c

and I was right before with my negative :L

cos(u) = (1-sin(u)^(2))^(1/2)
=
(1-x^(2))^(1/2)

Answer 27

the final answer is

arcsin^(2)(x)x + 2(1-x^(2))^(1/2)arcsin(x) - 2x + c

Answer 28

Going to check with wolfram alpha now :)

Answer 29

Confirmed :)
https://www.wolframalpha.com/input/?i=integrate+arcsin%28x%29^%282%29

Did you follow sorry it got sketchy in the integration by parts area

Answer 30

Hope this was helpful or you just wanted to see someone prove this :)

Answer 31

no..i was really asking someone to solve it hahaha =)))) you just use by parts and variable subs right? no trig sub?

Answer 32

well you are kind of doing a trig sub but not really as it is pretty intuitive.

Answer 33

remember that

cos(u) = \[\sqrt{1-x^{2}}\]
sin(u) = x
and
u = arcsin(x)

Answer 34

you need to remember to sub those values into the end equation, I had this same question except it wasn't squared on my calc 1 final, I killed it :)

Answer 35

but yes these questions are easy just very long

Answer 36

I should really have gone one step at a time instead of skipping steps though

Answer 37

sorry if I wasn't any help

Answer 38

nahh..izfine haha