 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 x into tan x upon sec x plus tan x dx. Let us now begin with this solution. Let us denote this definite integral by i. So we have i is equal to integral limit from 0 to pi x tan x upon sec x plus tan x dx. This is equal to integral limit from 0 to pi x into, now we can write tan x as sin x upon cos x by sec x can be written as 1 upon cos x plus tan x can be written as sin x upon cos x dx. And we can write this as this is equal to integral limit from 0 to pi x sin x upon 1 plus sin x dx and we mark this as equation 1. Now we know that integral limit from 0 to a fx dx is equal to integral limit from 0 to a f of a minus x dx. So we can write this equation 1 as i is equal to integral limit from 0 to pi pi minus x into sin of pi minus x upon 1 plus sin of pi minus x dx. This is equal to integral limit from 0 to pi pi minus x into, now sin of 180 minus theta is equal to sin theta. So we get sin pi minus x is equal to sin x upon 1 plus sin x dx and we mark this as equation 2. Now adding 1 and 2 we get i plus i is equal to integral limit from 0 to pi x sin x upon 1 plus sin x dx plus integral limit from 0 to pi pi minus x sin x upon 1 plus sin x dx. This implies 2i is equal to integral limit from 0 to pi x sin x upon 1 plus sin x dx plus integral limit from 0 to pi pi into sin x upon 1 plus sin x dx minus integral limit from 0 to pi x sin x upon 1 plus sin x dx. Now the first and the third term on the right hand side will cancel out. So we get 2i is equal to integral limit from 0 to pi pi into sin x upon 1 plus sin x dx. This is equal to pi integral limit from 0 to pi. Now adding and subtracting 1 in numerator we get 1 plus sin x minus 1 upon 1 plus sin x dx and this is equal to pi integral limit from 0 to pi 1 plus sin x upon 1 plus sin x minus 1 upon 1 plus sin x dx right. Now here 1 plus sin x will cancel out in numerator and denominator. So we are left with this is equal to pi integral limit from 0 to pi 1 minus 1 upon 1 plus sin x dx. We can write this as this is equal to pi integral limit from 0 to pi dx minus pi integral limit from 0 to pi 1 upon 1 plus sin x dx this is equal to pi. Now integrating 1 we get x. So we have x and limit is from 0 to pi minus pi integral limit from 0 to pi. Now multiplying the numerator and denominator by 1 minus sin x we get 1 minus sin x upon 1 plus sin x into 1 minus sin x dx. This is equal to pi. Now putting the limits we get here pi minus 0 because upper limit is pi. So putting pi in place of x we get pi minus lower limit is 0. So putting 0 in place of x we get 0 minus pi integral limit from 0 to pi 1 minus sin x upon now we know that a plus b into a minus b is equal to a square minus b square. So we get 1 minus sin square x dx since a plus b into a minus b is equal to a square minus b square. This is equal to now pi into pi is pi square. So we get pi square minus pi into 0 is 0 minus pi integral limit from 0 to pi 1 minus sin x upon now we know that 1 minus sin square x is equal to cos square x. So we get cos square x and denominator dx. This is equal to pi square minus pi integral limit from 0 to pi. Now we can write 1 minus sin x upon cos square x as 1 upon cos square x minus sin x upon cos square x dx and this is equal to pi square minus pi integral limit from 0 to pi. Now we can write 1 upon cos square x as sin square x minus sin x upon now we can write cos square x as cos x into cos x write dx and this is equal to pi square minus pi integral limit from 0 to pi sin square x minus now sin x upon cos x is equal to tan x. So we get tan x into and 1 upon cos x is equal to sin x. So we get sin x dx this is equal to pi square minus pi into now integral of sin square x is equal to tan x. So we get tan x minus and integral of tan x into sin x is equal to sec x. So we get sec x and limit is from 0 to pi. This is equal to pi square minus pi into now putting the limits we get tan pi minus sec pi minus tan 0 minus sec 0 because putting upper limit pi in place of x we get tan pi minus sec pi minus and putting lower limit 0 we get tan 0 minus sec 0. This is equal to pi square minus pi into now we know that tan pi is equal to 0. So we get 0 minus and sec pi is equal to minus 1. So we get minus 1 minus tan 0 again is 0. So we get 0 minus and sec 0 is equal to 1. So we get 1 this is equal to pi square minus pi into now minus into minus will become plus. So we get plus 1 and here again minus into minus will become plus. So we get plus 1 and this is equal to pi square minus 1 plus 1 is 2. So 2 into pi is 2 pi. So we have 2i is equal to now taking out pi common we get pi into pi minus 2 and this implies i is equal to pi by 2 into pi minus 2. So we get i is equal to pi by 2 into pi minus 2. Thus we write our answer as pi by 2 into pi minus 2. Hope you have understood the solution. Take care and bye.