Question

How do you integrate `int arcsinx` by integration by parts method?

Answer 1

`intsin^(-1)xdx=xsin^(-1)+(1-x^2)^(1/2)+C`

Explanation:

IBP formula

`I=intcolor(red)(u)v'dx=intcolor(red)(u)v-vcolor(red)(u')dx`

we ahve

`intsin^(-1)xdx`

let`" "color(red)( u=sin^(-1)x=>u'=1/sqrt(1-x^2))`

`v'=1=>v=x`

`I=xcolor(red)(sin^(-1)x)-intx/color(red)(sqrt(1-x^2))dx`

now
`intx/color(red)(sqrt(1-x^2))dx=intx(1-x^2)^(-1/2)dx`

by inspection we have

`=-(1-x^2)^(1/2)`

this is left as an exercise for the reader to verify

finally we have

`intsin^(-1)dx=xsin^(-1)+(1-x^2)^(1/2)+C`