`∫x sin^-1 x dx`

Explanation

`" Let " I = int " x sin"^(-1) x dx "` lt brgt ` I= sin^(-1) x int x dx - int [(d)/(dx) (sin^(-1) x) int x dx ] dx`
` =sin^(-1) x. (x^(2))/(2)- int ((1)/(sqrt(1-x^(2))).(x^(2))/(2))dx`
`=(x^(2))/(2).sin^(-1) x+ int ((1-1-x^(2))/(sqrt(1-x^(2))).(1)/(2))dx`
`=(x^(2))/(2).sin^(-1) x-(1)/(2) int (1)/(sqrt(1-x^(2)))dx`
`+(1)/(2) int (1-x^(2))/(sqrt(1-x^(2)))dx`
` =(x^(2))/(2).sin^(-1) x-(1)/(2) sin^(-1) x+(1)/(2) int sqrt(1-x^(2))dx`
` rArr I=(x^(2))/(2).sin^(-1) x-(1)/(2) sin^(-1) x`
` +(1)/(2) ((1)/(2) xsqrt(1-x^(2))+(1)/(2) sin^(-1) x)+C`
` rArr I = sin^(-1) x ((x^(2))/(2)-(1)/(4))+(1)/(4) xsqrt(1-x^(2))+C`
`rArr I=(sin^(-1))/(4) (2x^(2) -1)+(1)/(4) x sqrt(1-x^(2))+C`