 Hi and welcome to the session. I am Purva and I will help you with the following question. Evaluate the definite integral. Integral limit is from 0 to pi by 2 sin 2x into tan inverse sin x dx. Now we begin with the solution. Let us denote this definite integral by i. So we have i is equal to integral limit is from 0 to pi by 2 sin 2x into tan inverse sin x dx. And this is equal to integral limit is from 0 to pi by 2. Now we can write sin 2x as 2 sin x cos x. So we have sin 2x is equal to 2 into sin x into cos x into tan inverse sin x dx. Since we have sin 2 theta is equal to 2 sin theta into cos theta. Now let us put sin x is equal to 2 t. So we have put sin x is equal to t. Now differentiating we get, differentiating sin x we get cos x dx is equal to differentiating t we get dt. So we get cos x dx is equal to dt. Now when x is equal to lower limit that is 0 we have t is equal to, now putting here 0 in place of x we get sin of 0 is equal to 0. So we get t is equal to 0. And when x is equal to upper limit that is pi by 2 we have t is equal to. Now putting here pi by 2 in place of x we get sin pi by 2 is equal to 1. So we get t is equal to 1. So we get the limits as when x is equal to 0 we have t is equal to 0 and when x is equal to pi by 2 we have t is equal to 1. Putting all these values in i we get i is equal to integral limit is from 0 to 1 2 into t into tan inverse t dt. Or we can write this as this is equal to integral limit is from 0 to 1 t into tan inverse t dt. Now taking tan inverse t as the first function and t as the second function and applying Bipath's method we get this is equal to 2 into now first that is tan inverse t into integration of second. Now integration of t gives us t square upon 2 minus integration derivative of first that is derivative of tan inverse t gives us 1 upon 1 plus t square into integration of second that is t. So we have integration of t is t square by 2 dt and limit is from 0 to 1. And this is equal to 2 into now putting the limits we get tan inverse 1 into 1 by 2 minus tan inverse 0 into 0 because the upper limit is 1. So putting the upper limit here we get tan inverse 1 into 1 by 2 minus now putting the lower limit we get tan inverse 0 into 0 minus integral limit is from 0 to 1 t square upon 2 into 1 plus t square dt. And this is equal to 2 into 1 upon 2 into tan inverse 1 minus 2 into 1 upon 2 integral limit from 0 to 1 t square upon 1 plus t square dt. Opening the bracket we get here 2 into 1 upon 2 into tan inverse 1 minus 0 because this becomes 0 minus 2 into 1 upon 2 integral limit from 0 to 1 t square upon 1 plus t square dt. This is equal to now canceling 2 in numerator and denominator here and here we get tan inverse 1 minus integral limit from 0 to 1. Now we can write t square upon 1 plus t square as 1 minus 1 upon 1 plus t square dt. And this is equal to tan inverse 1 minus now we can write this whole thing as integral limit from 0 to 1 dt plus integral limit from 0 to 1 1 upon 1 plus t square dt right and this is equal to tan inverse 1 minus now integrating 1 we get t and limit is from 0 to 1 plus integrating 1 upon 1 plus t square we get tan inverse t and again the limit is from 0 to 1. This is equal to tan inverse 1 minus now putting the upper and lower limits here we get 1 minus 0 because the upper limit is 1 so putting 1 in place of t we get 1 and the lower limit is 0 so putting 0 in place of t we get 0 plus again putting the upper and lower limits we get tan inverse 1 minus tan inverse 0 and this is equal to tan inverse 1 minus 1 plus tan inverse 1 and tan inverse 0 is 0 so we get here minus 0 and this is equal to now tan inverse 1 is equal to pi by 4 so we get pi by 4 minus 1 plus again tan inverse 1 is equal to pi by 4 and we get this is equal to 2 pi by 4 minus 1 because pi by 4 plus pi by 4 gives 2 pi by 4 and minus 1 now cancelling 2 in numerator and denominator we get this is equal to pi by 2 minus 1 so we get our answer as pi by 2 minus 1 hence we write our answer as pi by 2 minus 1 hope you have understood the solution take care and bye