 Hello and welcome to the session. In this session we discuss the following question that says evaluate integral pi by 6 to pi by 3 sin x plus cos x upon square root of sin 2x dx. We know that integral of dx upon square root of 1 minus x square is equal to sin inverse x. This is the key idea that we use in this question. Let's now move on to the solution. We need to evaluate this integral. So we suppose I will be equal to the integral pi by 6 to pi by 3 sin x plus cos x upon square root of sin 2x dx. Suppose that sin x minus cos x be equal to t. So this means on differentiating both the sides we get cos x plus sin x dx would be equal to dt or sin x plus cos x dx is equal to dt. Now as in the integral we have the limits as pi by 6 to pi by 3 that is x goes from pi by 6 to pi by 3. Now when we have x equal to pi by 6 t would be equal to sin pi by 6 minus cos pi by 6 which means t would be equal to 1 by 2 minus root 3 by 2 which is equal to 1 minus root 3 whole upon 2. Now when we put x equal to pi by 3 we get t equal to sin pi by 3 minus cos pi by 3 that is we have t equal to root 3 by 2 minus 1 by 2 equal to root 3 minus 1 this whole upon 2. Also x minus cos x equal to t so this means sin x minus cos x the whole square is equal to t square that is we have sin square x cos square x minus 2 sin x into cos x is equal to t square. Now sin square x plus cos square x is 1 so 1 minus is equal to t square or you can say 1 minus sin 2x sin 2x is equal to t square which means that is equal to 1 minus we can rewrite the given integral i. i is equal to integral of limits where pi by 6 to pi by 3 so now the limits would be 1 minus root 3 by 2 to 3 minus 1 by 2 is equal to integral root 3 whole upon 2 root 3 minus 1 whole upon 2 and we have got sin x plus cos x dx and dt sin 2x sin square root of 1 minus t square. My integral of dx upon square root of 1 minus x square is equal to sin inverse x a is equal to n inverse t and the limit goes from root 3 whole upon 2 to root 3 minus 1 whole upon 2 so this means we have i is equal to sin inverse 9 place of t first we put this limit that is root 3 minus 1 whole upon 2 minus or sin inverse 9 place of t we will put this limit that is 1 minus root 3 whole upon 2. This gives us i is equal to sin inverse of root 3 minus 1 whole upon 2 minus sin inverse of minus of root 3 minus 1 whole upon 2 we know that sin inverse of minus x is equal to minus of sin inverse x so using this we get i is equal to sin inverse of root 3 minus 1 whole upon 2 plus plus of root 3 minus 1 whole upon 2 so this gives us i is equal to 2 into sin inverse of root 3 minus 1 whole upon 2 therefore we can say that we give an integral pi by 6 to pi by 3 sin x plus cos x this whole upon square root of dx is equal to 2 into sin inverse of root 3 minus 1 upon 2. So this is our final answer this can be the session we have understood the solution of this question.