 Hi and welcome to the session. I am Purva and I will help you with the following question. Proof the following integral, limit is from 0 to 1, sin inverse x dx is equal to pi by 2 minus 1. Now we begin with the solution. We denote this left hand side by i. So we have i is equal to integral, limit is from 0 to 1, sin inverse x dx. Now we put x is equal to sin t. So put x is equal to sin t. Now differentiating we get dx is equal to cos t dt. Now when x is equal to lower limit that is when x is equal to 0 then t is equal to sin inverse 0 right and we know that sin inverse 0 is equal to 0. So when x is equal to 0 t is also equal to 0. And when x is equal to 1 that is upper limit then we have t is equal to sin inverse 1. Now we know that sin inverse 1 is equal to pi by 2. So we get t is equal to pi by 2. Now putting all these values in i we get i is equal to integral limit is from 0 to pi by 2 sin inverse sin t into cos t dt. And we can write this as this is equal to integral limit is from 0 to pi by 2. Now sin inverse sin t is equal to t into cos t dt. Taking this t as the first term and cos t as the second term and applying Bipart's method we get this is equal to first that is t into integration of second that is integration of cos t which is equal to sin t limit is from 0 to pi by 2 minus integration derivative of first. Now derivative of t is 1 into integration of cos t that is sin t dt. And again here limit is from 0 to pi by 2. Now we put the limits in place of t. So we have this is equal to upper limit that is pi by 2 into sin pi by 2 minus lower limit that is 0. So this whole thing becomes equal to 0 minus now integral sin t is equal to minus cos t. And here again the limit is from 0 to pi by 2. And this is further equal to pi by 2 into now sin pi by 2 is equal to 1. So we get 1 and here minus into minus becomes plus and now we put the limits. So we get cos pi by 2 minus cos 0 right. And this is equal to pi by 2 now cos pi by 2 is 0. So we don't write it minus now cos 0 is 1. So we get our answer as pi by 2 minus 1 which is equal to the right hand side. Thus we have proved that integral limit from 0 to 1 sin inverse x dx is equal to pi by 2 minus 1. Hope you have understood the solution. Take care and God bless you.