 Hello and how are you all today? The question says evaluate i is equal to integral 0 to pi by 2 sin square x upon sin x plus cos x dx. So let's proceed with the solution. We have been given the integral, the definite integral as integral 0 to pi by 2 sin square x upon sin x plus cos x dx. 0 to pi by 2. Now here we can write x pi by 2 minus x for each and every case. Further it will become i is equal to 0 to pi by 2 sin square pi by 2 minus x is cos square x upon sin pi by 2 minus x is cos x. Now this will become sin x. Let this be the second equation, this be the first equation. Now on adding the first and the second equation we get 2i equal to integral 0 to pi by 2. Now here the denominator is same for the case. So we have sin square x plus cos square x upon sin x plus cos x dx, right? So we have 2i equal to integral 0 to pi by 2 sin square x plus cos square x is equal to 1, right? So we have 1. Now let us divide and multiply our denominator by root 2. So here we can write sin x upon root 2 plus cos x upon root 2 dx. For 2i is equal to integral 0 to pi by 2 1 upon, let us take 1 by root 2, comment out. Now here if a will be x and b will be 1 by, let me rewrite it. Sin x the value of 1 by root 2 can be written as cos pi by 4, isn't it? We can write 1 by root 2 which is over here as cos pi by 4 and similarly x into 1 by root 2 can be written as sin pi by 4. So in doing so we have i equal to 1 by root 2 integral 0 to pi by 2. Now here sin a cos b plus cos a sin b is the expansion of sin a plus b, isn't it? Where a is x and b is pi by 4. So we have this. Now proceeding it we have 2i equal to 1 by root integral 0 to pi by 2. 1 upon sin a plus b can be written as pi by 4 was over here. So x plus pi by 4 dx, right? So we know that the formula for integral cos a theta is log mod cos a theta which is x plus pi by 4 over here minus cot theta that is x plus pi by 4 0 to pi by 0 to pi by. So we have 2i equal to 1 by root 2 log mod cos a pi by 2 plus pi by 4 by 4 plus tan pi by 4 mod close minus log mod cos a pi by 4 minus cot pi by 4 mod close bracket close. So on writing down its values we have 1 by root 2 log mod root 2 plus 1 minus log mod root 2 minus 1 which is equal to 2i is equal to 1 by root 2 log mod root 2 plus 1 upon root 2 minus 1 plus c. Sorry, here this was a definite integral question. 2i equal to 1 by root 2 on rationalizing we have the denominator that is root 2 plus 1 into root 2 minus 1 upon root 2 minus 1 sorry plus root 2 whole square minus 1 the whole square. 1 by root 2 log 2 plus 1 the whole square upon 2 minus 1 that is 1 which is or i is equal to 1 by root 2 log mod root 2 plus 1. So this is the required answer to the given question hope you understood it well.