 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 9 plus 16 sin 2x where limit is from 0 to pi by 4. Now we begin with the solution. We denote this integral by i. So we have i is equal to integral limit is from 0 to pi by 4 sin x plus cos x upon 9 plus 16 sin 2x dx. This is equal to integral limit from 0 to pi by 4 sin x plus cos x upon 9 plus 16 into 1 minus sin x minus cos x whole square dx. Now I will tell you how we have got this second step. Now sin x minus cos x whole square can be written as this is equal to sin square x plus cos square x minus 2 into sin x into cos x right by applying the formula of a minus b whole square which is equal to a square plus b square minus 2 into a into b. Now this is further equal to now we know that sin square x plus cos square x is equal to 1. So we get this is equal to 1 minus and 2 sin x cos x is equal to sin 2x so sin 2x. So we have got sin x minus cos x whole square is equal to 1 minus sin 2x. We can also write this as this implies sin 2x is equal to 1 minus sin x minus cos x whole square. So in this first step we replace sin 2x by 1 minus sin x minus cos x whole square to get this second step. So you have understood how we have got the second step. This is equal to integral limit from 0 to pi by 4 sin x plus cos x upon 9 plus 16 minus 16 into sin x minus cos x whole square dx right which is further equal to integral limit from 0 to pi by 4 sin x plus cos x upon 9 plus 16 is equal to 25 minus 16 into sin x minus cos x whole square dx. Now we put sin x minus cos x equal to t. So put sin x minus cos x equal to t. Now we differentiate it. So differentiating we get after differentiation sin x gives cos x. So cos x and after differentiation minus cos x is equal to sin x so plus sin x dx equal to dt. Now when x is equal to lower limit that is when x is equal to 0 then t is equal to putting here 0 in place of x we get sin 0 minus cos 0 that is this is equal to now sin 0 is equal to 0 minus cos 0 is equal to 1 which is further equal to 0 minus 1 is minus 1. And when x is equal to upper limit that is when x is equal to pi by 4 we have t is equal to sin pi by 4 minus cos pi by 4 which is equal to now sin pi by 4 is 0 and cos pi by 4 is also 0. So we get this is equal to 0. So when x is equal to 0 we have t is equal to minus 1 and when x is equal to pi by 4 we have t is equal to 0. Putting all these values in i we get i is equal to integral limit from minus 1 to 0 dt upon 25 minus 16 t square. Now multiplying and dividing the denominator by 16 we get this is equal to 1 by 16 integral limit from minus 1 to 0 dt upon 25 by 16 minus dividing here by 16 we get only t square so t square. This is equal to 1 by 16 integral limit from minus 1 to 0 dt upon now we can write 25 by 16 as 5 by 4 whole square minus t square this is equal to 1 by 16. Now we know that integral dx upon a square minus x square is equal to 1 by 2 a into log a plus x upon a minus x we know this formula. So by applying this formula here we get 1 by 2 into a that is 5 by 4 multiplied by log a plus x that is 5 by 4 plus t upon 5 by 4 minus t and limit is from minus 1 to 0. Canceling the common factor 2 we get here 2 so further this is equal to 1 by 16 into we can write 1 upon 5 by 2 as 2 by 5 into log 5 by 4 plus t upon 5 by 4 minus t limit is from minus 1 to 0. Again we cancel the common factor 2 from numerator and denominator we get here 8 so this is equal to now 1 upon 8 into 5 is 40 into log 5 by 4 plus now the upper limit is 0 so we put 0 in place of t upon 5 by 4 minus 0 minus log 5 by 4 plus now the lower limit is minus 1 so we put minus 1 in place of t upon 5 by 4 minus minus 1 this is equal to 1 by 40 into log 5 by 4 plus 0 is 5 by 4 upon 5 by 4 minus 0 is 5 by 4 minus log 5 by 4 minus 1 upon 5 by 4 minus into minus is plus plus 1 this is equal to 1 by 40 into log now 5 by 4 upon 5 by 4 is 1 minus log 5 by 4 minus 1 is 1 by 4 upon 5 by 4 plus 1 is 9 by 4 this is equal to 1 by 40 into log 1 minus log now we cancel out 4 and 4 we are left with 1 by 9 so we get 1 by 9 and this is further equal to 1 by 40 into log 1 minus now we can write log 1 by 9 as log 1 minus log 9 and this is equal to 1 by 40 into log 1 minus log 1 plus log 9 right now cancelling out plus log 1 and minus log 1 we are left with this is equal to 1 by 40 into log 9 so we get our answer as 1 by 40 into log 9 this is our answer hope you have understood the solution take care and have a nice day