 Hi and welcome to the session. Let's discuss the following question. The question says integrate the following functions sin inverse root x minus cos inverse root x by sin inverse root x plus cos inverse root x. Let's now begin with the solution. In this question we have to find integral of sin inverse root x minus cos inverse root x by sin inverse root x plus cos inverse root x with respect to x. First of all we should know that sin inverse root x plus cos inverse root x is equal to pi by 2. And sin inverse root x plus cos inverse root x equals to pi by 2 implies cos inverse root x is equal to pi by 2 minus sin inverse root x. By substituting pi by 2 in place of sin inverse root x plus cos inverse root x and pi by 2 minus sin inverse root x in place of cos inverse root x, we get integral sin inverse root x minus pi by 2 plus sin inverse root x by pi by 2 dx. This is equal to 2 pi pi into integral 2 sin inverse root x minus pi by 2 dx. This is equal to 4 pi pi into integral sin inverse root x dx minus integral dx. Now let i1 is equal to integral of sin inverse root x with respect to x. Now put root x equals to t. Now root x equals to t implies x is equal to t square and this implies dx is equal to 2t into dt. So by substituting t in place of root x and 2t dt in place of dx, we get i1 as integral into sin inverse t into dt. This is equal to 2 into integral of sin inverse t into t with respect to t. Now here we have product of two functions and we know that if u and v are two functions of t, then integral of uv with respect to t is equal to u into integral of v with respect to t minus integral derivative of u with respect to t into integral of v with respect to t whole with respect to t. We choose the first function as the function which comes first in the word iLate. Here we have algebraic function and trigonometric inverse function. According to iLate, we choose first function as inverse trigonometric function that means sin inverse t. So using this rule, this integral is equal to sin inverse t into integral of t with respect to t minus derivative of sin inverse t into integral of t with respect to t whole dt. This is equal to 2 into sin inverse t. We know that integral of x to the power n with respect to x is x to the power n plus 1 pi n plus 1 plus c. So using this formula integral of t with respect to t is t square by 2 minus integral derivative of sin inverse t is 1 by square root of 1 minus t square into integral of t dt is t square by 2 whole dt. Now this is equal to t square sin inverse t minus 2 into 1 by 2 into integral t square pi square root of 1 minus t square with respect to t. And this is equal to t square sin inverse t minus integral t square pi square root of 1 minus t square dt. This is equal to t square sin inverse integral we are now going to add and subtract 1 in the numerator. So we have t square minus 1 plus 1 by square root of 1 minus t square dt. This is equal to t square sin inverse integral t square minus 1 plus integral 1 by plus t taking minus common from the numerator we get plus plus integral. So using this formula of these integrals I am going to replace t by root of 1 by 2 sin inverse integral t to sin inverse root x plus root x into square root of 1 minus x by root of 1 minus t square dt. So by substituting values, we get 2 can be written as 2x minus 1 by 2 into sin inverse root x root 1 minus x by 2 plus c1 minus integral dx is x plus c2 is equal to minus 1 by sin inverse root x plus into root inverse root x 2 into root x into root 1 minus square root x minus 1 by 2 into sin inverse root x minus 1 by 2 into sin inverse root x