Seriously need help.
Integration by part..sorry i don't know how to use Latex ,,,so i am writing like this..
let,I =∫x sin^-1x dx/√1-x^2
=x sin^-1x∫1/√1-x^2 dx-∫{d/dx x sin^-1x ∫1/√1-x^2 dx}dx
=x sin^-1x .sin^-1x-∫x sin^-1x/√1-x^2 dx
Since ,I=∫x sin^-1x dx/√1-x^2
=x sin^-2x-I
so,I+I=x sin^-2x
=2I=x sin^-2x
=I=x sin^-2x/2 +C
is this right?

Question

Seriously need help.
Integration by part..sorry i don't know how to use Latex ,,,so i am writing like this..
let,I =∫x sin^-1x dx/√1-x^2
      =x sin^-1x∫1/√1-x^2 dx-∫{d/dx x sin^-1x ∫1/√1-x^2 dx}dx
=x sin^-1x .sin^-1x-∫x sin^-1x/√1-x^2 dx
Since ,I=∫x sin^-1x dx/√1-x^2
=x sin^-2x-I
so,I+I=x sin^-2x
=>2I=x sin^-2x
=>I=x sin^-2x/2 +C
is this right?

inkyvoyd · Answer

hint: click the equation button.

zepdrix · Answer

Ya sorry I can't read your notes very easily.
Your solution doesn't look quite right.

I'll draw a picture for you, to show what you want your parts to look like.

zepdrix · Answer

|dw:1392014075811:dw|I introduced a negative twice, hopefully that isn't too confusing.