 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says integrate the following function that is x sin inverse x. So let us start with the solution to this question. Now we have to find i that is integral x sin inverse x dx. Now we see that sin inverse x becomes the first function and x will be the second function according to the i-late rule because inverse function is given preference over the algebraic function. So we see that i will be equal to integration this and by applying integration bypass as we have seen in the earlier questions we get sin inverse x into integral x dx minus integral d by dx of sin inverse x integral of x dx this will be equal to sin inverse x into x square by 2 because integral x dx is x square by 2 minus integral now d by dx of sin inverse x is 1 upon under the root 1 minus x square into x square by 2 we get from integral x dx. Now this is equal to x square by 2 sin inverse x minus 1 by 2 integral x square by under the root 1 minus x square dx. Now we call this as i1. So we have x square by 2 sin inverse x minus 1 by 2 i1. So now we have to find i1 where i1 is integral x square by under the root 1 minus x square dx. Now what we do is we put x equal to sin theta if x equals to sin theta then dx will be equal to cos theta d theta therefore i1 will be equal to integral x square that is sin square theta divided by under the root 1 minus sin square theta into cos theta d theta this is equal to integral sin square theta by cos theta into cos theta d theta because square root of 1 minus sin square theta is cos theta cos theta gets cancelled with cos theta and we have integral sin square theta d theta. Now this can be further written as 1 by 2 integral 1 minus cos 2 theta d theta because sin square theta is equal to 1 minus cos 2 theta divided by 2 this is equal to 1 by 2 integral d theta minus 1 by 2 integral cos 2 theta d theta we have simply separated the terms that will be equal to half of theta minus half of sin theta cos theta plus some constant c1. Now let us see how do we get this from this this happens because we see that integral cos 2 theta will be equal to sin 2 theta divided by 2 plus some constant c1 now here we have integral cos 2 theta d theta now we know that sin 2 theta is same as 2 sin theta cos theta this divided by 2 plus c1 2 gets cancelled with 2 and we are left with sin theta cos theta plus c1 so we have this now we put back the value of theta from here we can see that theta is equal to sin inverse x so we can write this as 1 by 2 sin inverse x minus 1 by 2 into x into square root of 1 minus x square plus the constant c1 because sin theta is x so this implies cos theta will be equal to under the root 1 minus sin square theta that is under the root 1 minus x square therefore i becomes x square by 2 sin inverse x minus 1 by 2 into 1 by 2 sin inverse x minus 1 by 2 x into square root of 1 minus x square plus the constant c1 now let us open the brackets on doing so we get x square by 2 sin inverse x minus 1 by 4 sin inverse x plus 1 by 4 x into square root of 1 minus x square plus c1 and this can be further written as 1 by 4 sin inverse x into 2 x square minus 1 plus x into square root of 1 minus x square divided by 4 plus we call this constant say c since we see that constant is unknown so we can call it c c1 anything so our answer to this question is 1 by 4 sin inverse x into 2 x square minus 1 plus x into square root of 1 minus x square divided by 4 plus c so this is our answer to the question i hope that you understood the question and enjoyed the session have a good day