Question

I is equal to integrate x Sin inverse x square divided by root of 1 - x square into DX​

Answer 1

Answer:

if you get the answer please send it to me

I tried but failed to get it

Answer 2

Answer:

Step-by-step explanation:

∫x*arcsinx  dx / root 1-x^2

arcsinx = t

dx/root 1-x^2 = dt

arcsinx=t

x=sin t

∫sint t dt

applying integration by parts

t∫sin t dt - ∫(∫sin t dt ) dt

-tcost+sint+c

-arcsinx ( cos(sin^- x ) + sin(arc sin x)+c

-arcsinx( root (1-x^2) ) +x+C

arcsinx means sin inverse x