 Hi and how are you all today? The question says evaluate integral sin inverse x upon x square dx. So here we need to integrate sin inverse x upon x square with respect to x. Here we can write this as x raised to the power minus 2 sin inverse x dx right. Now integrating by parts we have let this be the first function and this be the second function. So it is first function into integral of second function minus integral derivative of the first function into integral of second function dx. Sin inverse x into, now here integral of this function will be x raised to the power minus 2 plus 1 upon minus 2 plus 1 minus integral derivative of sin inverse x is equal to 1 upon under root 1 minus x square right into integral of this function is x raised to the power minus 2 plus 1 divided by minus 2 plus 1 into, now this term is minus sin inverse x upon x minus integral 1 upon x under root. Now converting it in the form of a square minus x square it will be 1 square minus x square minus sin inverse x upon x minus we know that we can write integral dx upon x under root a square minus x square equal to log a upon x minus under root 1 minus x square upon x they both are into mod plus c right. So here a is equal to 1 so we will have this was minus 1 so minus minus will become plus over here so further we have plus log minus under root 1 minus x square also be written as minus under root 1 minus x square of this question hope you understood it well and enjoyed it to have a nice day.