 Hello and welcome to the session I am Deepika here. Let's discuss the question which says evaluate the following integrals using substitution. Integral from 0 to 1 sign inverse 2 is upon 1 plus x square tx. So let's start the solution. Let x is equal to tan theta then dx is equal to secant square theta d theta. So the new limits are when x is equal to 0 this implies tan theta is equal to 0 and this implies theta is equal to 0. And when t is equal to 1 this implies tan theta is equal to 1 and this implies theta is equal to pi by 4. So therefore integral from 0 to 1 sign inverse x upon 1 plus x square dx is equal to integral from 0 to pi by 4 sign inverse 2 tan theta over 1 plus tan square theta into secant square theta d theta. This is equal to integral from 0 to pi by 4 sign inverse. Now this is a formula of sin 2 theta into secant square theta d theta and this is equal to integral from 0 to pi by 4 2 theta into secant square theta d theta. Now let us integrate this function by parts. Let us take theta as the first function and secant square theta as the second function. So this is equal to into now first function into integral of the second function that is secant square theta d theta minus integral derivative of the first function that is 1 into integral of the second function. Now this is equal to into theta into integral of secant square theta d theta is tan theta minus integral of 1 into secant square theta d theta and the integral of secant square theta d theta is tan theta and this is equal to into theta tan theta minus log of mod sec theta. Therefore integral from 0 to 1 sign inverse 2 x upon 1 plus x square d x is equal to theta tan theta minus 2 log mod of sec theta from 0 to pi by 4 this is equal to 2 pi by 4 tan pi by 4 minus 2 log pi by 4 minus into 0 into tan 0 that is 0 minus 2 log secant 0. Now this is equal to pi by 2 into tan pi by 4 now tan pi by 4 is 1 so pi by 2 into 1 is pi by 2 minus tan pi by 4 now secant pi by 4 is root 2 so this is minus 2 log root 2 minus 0 plus 2 log secant 0 now secant 0 is 1 so this is 2 log 1 this is equal to pi by 2 minus 2 into log root 2 minus log 1 this is again equal to pi by 2 because log m over n is equal to log m minus log n and this is again equal to pi by 2 minus log because log m raise to power n is equal to n log m and this is equal to pi by 2 minus log 2. Hence the answer further above question is pi by 2 minus log 2 I hope the solution is clear to you why and take care.