 Hi and welcome to the session. I am Purva and I will help you with the following question. Evaluate the definite integral sin x plus cos x upon under root sin 2x where limit is from pi by 6 to pi by 3. Now we begin with the solution. Now we denote this integral by i. So we have i is equal to integral limit is from pi by 6 to pi by 3 sin x plus cos x upon under root sin 2x dx. This is equal to integral limit from pi by 6 to pi by 3 sin x plus cos x upon under root 1 minus cos x minus sin x whole square dx. Now I will tell you how we have got this second step. Now 1 minus cos x minus sin x whole square can be written as this is equal to 1 minus a minus b whole square is equal to a square plus b square minus 2ab. So we get cos square x plus sin square x minus 2 sin x into cos x and this is equal to 1 minus. Now cos square x plus sin square x is equal to 1 so we get 1 minus 2 sin x into cos x is equal to sin 2x. So we get sin 2x and cancelling out plus 1 and minus 1 we get this is equal to minus into minuses plus so plus sin 2x. So we have got sin 2x is equal to 1 minus cos x minus sin x whole square. So you have understood how we have got the second step. Now we put cos x minus sin x equal to t. So put cos x minus sin x equal to t. Now we differentiate it. So differentiating we get after differentiation cos x gives minus sin x and minus sin x gives minus cos x dx which is equal to dt because after differentiation t gives dt or we can write this as minus of sin x plus cos x dx is equal to dt when x is equal to lower limit that is when x is equal to pi by 6 we have t is equal to cos. Now putting x is equal to pi by 6 here we get cos pi by 6 minus sin pi by 6 which is further equal to now we know that cos pi by 6 is equal to root 3 by 2 minus and sin pi by 6 is equal to 1 by 2. So this is equal to root 3 minus 1 upon 2 and when x is equal to upper limit that is when x is equal to pi by 3 we have t is equal to cos pi by 3 minus sin pi by 3 which is further equal to now cos pi by 3 is equal to 1 by 2 minus sin pi by 3 is equal to root 3 by 2. So this is equal to 1 minus root 3 upon 2. Putting all these values in I we get I is equal to minus integral limit is from root 3 minus 1 by 2 to 1 minus root 3 by 2 1 upon under root 1 minus t square dt. This is equal to now we know that integral dx upon a square minus x square is equal to sin inverse x by a. So we have here minus sin inverse here x is equal to t so t by and a is equal to 1 so 1 and limit is from root 3 minus 1 by 2 to 1 minus root 3 by 2. This is further equal to minus sin inverse upper limit that is 1 minus root 3 by 2 minus sin inverse lower limit that is root 3 minus 1 by 2. This is equal to minus sin inverse 1 minus root 3 by 2 and minus into minuses plus sin inverse root 3 minus 1 by 2. And this is further equal to we can write this as sin inverse root 3 minus 1 by 2 because we know that sin inverse minus x is equal to minus sin inverse x so minus sin inverse 1 minus root 3 by 2 gives us sin inverse root 3 minus 1 by 2 plus sin inverse root 3 minus 1 by 2 and this is equal to 2 sin inverse root 3 minus 1 by 2. Thus we get the answer as 2 sin inverse root 3 minus 1 by 2. So this is the answer of the question. Hope you have understood the solution. Take care and God bless you.