 Hi, and welcome to the session. I am Purva and I will help you with the following question. Evaluate the definite integral, integral limit from 0 to pi by 4 sin x into cos x upon cos raised to the power 4x plus sin raised to the power 4x dx. Let us begin with the solution. Let us denote this definite integral by i. So we have i is equal to integral limit from 0 to pi by 4 sin x into cos x upon cos raised to the power 4x plus sin raised to the power 4x dx. Now dividing the numerator and denominator by cos raised to the power 4x we get this is equal to integral limit from 0 to pi by 4 sin x into cos x upon cos raised to the power 4x upon cos raised to the power 4x divided by cos raised to the power 4x plus sin raised to the power 4x divided by cos raised to the power 4x dx. And this is equal to integral limit from 0 to pi by 4. Now we can write the numerator as tan x into 6 square x because 1 cos x will cancel here and we will get sin x upon cos cube x which we can write as tan x into 6 square x upon, now we write the denominator as 1 plus tan raised to the power 4x dx. Now we put tan square x equal to t. So put tan square x equal to t and differentiating this we get, differentiating tan square x we get 2 into tan x into 6 square x dx is equal to, now differentiating t we get dt or we can write this as tan x into 6 square x dx is equal to dt by 2. Now when x is equal to lower limit that is x is equal to 0 we have t is equal to, now putting here 0 in place of x we get t is equal to tan square 0 that is 0. So we get when x is equal to 0 t is equal to 0 and when x is equal to upper limit that is pi by 4 we have t is equal to now putting here pi by 4 in place of x we get t is equal to tan square pi by 4 which is equal to 1. So we get when x is equal to pi by 4 we have t is equal to 1. Now putting all these values in i we get i is equal to 1 upon 2 integral limit from 0 to 1 1 upon 1 plus t square dt and we get this is equal to 1 upon 2 into now integrating 1 upon 1 plus t square we get tan inverse t and the limit is from 0 to 1 and this is equal to 1 upon 2 into now putting the limits we get tan inverse 1 minus tan inverse 0 because we know that the upper limit is 1. So putting the value of 1 in place of t we get tan inverse 1 minus lower limit is 0. So putting 0 in place of t we get tan inverse 0 and this is further equal to 1 upon 2 into now tan inverse 1 is equal to pi by 4. So we get pi by 4 minus now tan inverse 0 is 0. So we get 0 hence we get this is equal to 1 by 2 into pi by 4 which gives us pi by 8. So we get our answer as pi by 8. So we write our answer as pi by 8. Hope you have understood the solution. Bye and take care.