 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 2 cos square x upon cos square x plus 4 sin square x dx. Let us now 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 2 cos square x upon cos square x plus 4 sin square x dx. This is equal to integral limit from 0 to pi by 2 cos square x upon cos square x plus 4. Now we can write sin square x as 1 minus cos square x dx. This is equal to integral limit from 0 to pi by 2 cos square x upon cos square x plus, now open the bracket we get 4 minus 4 cos square x dx and this is further equal to integral limit from 0 to pi by 2 cos square x upon 4 minus 3 cos square x dx because cos square x minus 4 cos square x gives us minus 3 cos square x. This is equal to now multiplying the numerator and denominator by minus 3 we get minus 1 upon 3 integral limit from 0 to pi by 2 minus 3 cos square x upon 4 minus 3 cos square x dx. This is equal to minus 1 upon 3 integral limit from 0 to pi by 2. Now adding and subtracting 4 in numerator we get 4 minus 3 cos square x minus 4 upon 4 minus 3 cos square x dx and we can write this as this is equal to minus 1 upon 3 integral limit from 0 to pi by 2 4 minus 3 cos square x upon 4 minus 3 cos square x minus 4 upon 4 minus 3 cos square x dx. This is equal to minus 1 upon 3 integral limit from 0 to pi by 2. Now canceling numerator and denominator here we get 1 minus 4 upon 4 minus 3 cos square x dx and we can write this as this is equal to minus 1 upon 3 integral limit from 0 to pi by 2 dx minus into minus will become plus so plus 4 upon 3 integral limit from 0 to pi by 2 dx upon 4 minus 3 cos square x. This is equal to minus 1 upon 3 now integrating 1 we get x and limit is from 0 to pi by 2 plus 4 upon 3 integral limit from 0 to pi by 2 dx upon 4 minus 3 cos square x and this is equal to minus 1 upon 3 now putting the limits we get pi by 2 minus 0 because the upper limit is pi by 2 so putting pi by 2 in place of x we get pi by 2 and lower limit is 0 putting 0 in place of x we get 0 plus 4 upon 3 integral limit from 0 to pi by 2 dx upon 4 minus 3 cos square x and this is equal to minus pi by 6 because pi by 2 minus 0 is equal to pi by 2 and pi by 2 into minus 1 upon 3 will give us minus pi by 6 plus 4 upon 3 integral limit from 0 to pi by 2 dx upon 4 minus 3 cos square x. This is equal to minus pi upon 6 plus 4 by 3 integral limit from 0 to pi by 2 now we divide the numerator and denominator by cos square x so we get 1 upon cos square x upon 4 upon cos square x minus 3 cos square x upon cos square x dx and this is equal to minus pi by 6 plus 4 by 3 integral limit from 0 to pi by 2 now 1 upon cos square x is equal to sex square x so we get sex square x upon 4 sex square x minus 3 dx and this is equal to minus pi by 6 plus 4 by 3 integral limit from 0 to pi by 2 sex square x upon 4 into now we can write sex square x as 1 plus tan square x so 1 plus tan square x minus 3 dx this is equal to minus pi by 6 plus 4 by 3 integral limit from 0 to pi by 2 sex square x upon now opening the bracket we get 4 plus 4 tan square x minus 3 dx and this is equal to minus pi by 6 plus 4 by 3 integral limit from 0 to pi by 2 sex square x upon now 4 minus 3 is equal to 1 so 1 plus 4 tan square x dx this is equal to minus pi upon 6 plus now we denote 4 upon 3 integral limit from 0 to pi by 2 sex square x upon 1 plus 4 tan square x dx by i1 so we write plus i1 now we consider i1 so we have i1 is equal to 4 upon 3 integral limit from 0 to pi by 2 sex square x upon 1 plus 4 tan square x dx now we put tan x equal to t so put tan x equal to t differentiating we get sex square x dx is equal to dt now when x is equal to lower limit that is when x is equal to 0 we have t is equal to 0 because putting here 0 in place of x we get tan 0 which is equal to 0 so we get t is equal to 0 and when x is equal to upper limit that is when x is equal to pi by 2 we have t is equal to tan pi by 2 and tan pi by 2 is equal to infinity so we get t is equal to infinity putting all these values in i1 we get i1 is equal to 4 by 3 integral limit from 0 to infinity dt upon 1 plus 4 t square this is equal to now multiplying and dividing the denominator by 4 we get 4 upon 3 into 1 upon 4 integral limit from 0 to infinity dt upon 1 by 4 plus t square cancelling 4 numerator and denominator here we get this is equal to 1 by 3 integral limit from 0 to infinity dt upon now we can write 1 upon 4 as 1 upon 2 whole square plus t square now since we know that integral dx upon a square plus x square is equal to 1 upon a tan inverse x by a therefore we can write this as this is equal to 1 upon 3 into 1 upon 2 tan inverse t upon 1 by 2 and limit is from 0 to infinity so we apply this formula dx upon a square plus x square integral and this is equal to 1 upon a tan inverse x upon a so in this taking a is equal to 1 upon 2 and x is equal to t we get 1 upon 1 by 2 tan inverse t upon 1 by 2 and this is equal to 2 upon 3 tan inverse 2 t and limit is from 0 to infinity this is equal to 2 upon 3 now putting the limits we get tan inverse infinity minus tan inverse 0 because we have the upper limit as infinity so putting infinity in place of t we get tan inverse infinity minus a lower limit is 0 so putting 0 in place of t we get tan inverse 0 and this is equal to 2 by 3 into now we know that tan of pi by 2 is equal to infinity so we get tan inverse infinity is equal to pi by 2 minus and we know that tan of 0 is 0 so we get tan inverse 0 is equal to 0 and this is equal to pi by 3 so we get i1 is equal to pi by 3 therefore we have i is equal to minus pi by 6 plus pi by 3 and this is equal to pi by 6 so we get i is equal to pi by 6 thus we get our answer as pi by 6 hope you have understood the solution bye and take care