 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 cos inverse x. So let us see the solution to this question. We have to find integral of x cos inverse x dx. So first of all we determine which is the first function and which is the second function. Since cos inverse x is the inverse function so we know that as per the islet rule inverse function is given preference over algebraic function. So this will be the first function, this will be the second function. We call this i. So i will be equal to first function that is cos inverse x into integral of second function that is x dx minus integral of d by dx of first function that is cos inverse x into integral of second function that is x dx into dx. This is equal to cos inverse x into x square by 2 minus. Now we know that d by dx of cos inverse x is minus 1 by square root of 1 minus x square into integral of x dx is x square by 2 into dx. This will be equal to x square by 2 into cos inverse x minus sign comes out of integral we have plus 1 by 2 integral x square by square root of 1 minus x square dx. This is equal to x square by 2 cos inverse x plus 1 by 2 i 1 where i 1 is x square divided by square root of 1 minus x square dx. Now we put x equal to cos theta so dx will be equal to minus sin theta d theta. Therefore i 1 will be equal to integral cos square theta into minus sin theta divided by square root of 1 minus x square that is 1 minus cos square theta d theta. This will be equal to now we see that this is written as cos square theta minus sign comes out of the integral sign into sin theta. Now square root of 1 minus cos square theta is sin theta d theta. Sin theta gets cancelled with sin theta and we have minus integral cos square theta d theta. This is equal to minus 1 by 2 integral 1 plus cos 2 theta d theta. This happens because we see that cos square theta is equal to 1 plus cos 2 theta divided by 2 because 1 plus cos 2 theta is equal to 2 cos square theta. Now taking them separately we get minus 1 by 2 integral d theta minus 1 by 2 integral cos 2 theta d theta. This will be equal to minus 1 by 2 theta plus sin 2 theta divided by 2 plus a constant c. This will be equal to minus 1 by 2 theta plus 1 by 2 into 2 sin theta cos theta plus c. This is equal to minus 1 by 2 theta plus 2 gets cancelled with 2. We have theta plus cos theta into now sin theta we can write as square root of 1 minus cos square theta plus the constant c. This is equal to minus 1 by 2 theta plus cos inverse x plus x into square root of 1 minus x square plus constant c. Therefore we can say that i is equal to x square by 2 cos inverse x plus half of minus 1 by 2 into cos inverse x plus x into square root of 1 minus x square plus c. This is equal to minus 1 by 2 cos inverse x minus 1 by 4 cos inverse x minus 1 by 4 into x into square root of 1 minus x square plus c. So our answer to this question is cos inverse x by 4 into 2 x square minus 1 by 4 into 2 x square minus 1 minus x by 4 into square root of 1 minus x square plus c. This we get by taking cos inverse x by 4 common from these two terms. So we get 2 x square minus 1 in the bracket. This remains as it is. So I hope that you understood the question and enjoyed the session. Have a good day.