 I am welcome to your session. Let us discuss the following question. The question says, evaluate the following integral of log 1 plus cos x from 0 to pi. Let's now begin with the solution. Let i is equal to integral of log 1 plus cos x from 0 to pi. We know that 1 plus cos theta is equal to 2 cos square theta by 2. So using this formula, we get this integral as log 2 cos square x pi 2 lower limit is 0, half a limit is pi. Now let fx is equal to 2 into cos x pi 2 whole square. Now f pi minus x is equal to log minus x pi 2 whole square. We know that cos pi minus theta is minus cos theta. So this is equal to log 2 minus cos x pi 2 whole square. This is equal to log 2 cos square x pi 2. And this is equal to fx, saying property of definite integrals which says integral of fx from 0 to 2a is equal to 2 times integral of fx from 0 to a if a minus x is equal to fx. And it is equal to 0 to a minus x is equal to minus fx. We get i as integral of log 2 cos square x pi 2 0 to pi by 2 into 2 where a is equal to sorry 2 is equal to pi. So a is equal to pi by 2. Now by using the property of definite integrals which says integral of fx from 0 to a is equal to integral of f a minus x from 0 to a. We get i as integral of log cos square pi by 2 minus x pi 2 from 0 to pi by 2 into 2. This is equal to 2 into integral of log 2 sine square x by 2 from 0 to pi by 2. Let's name this equation as equation number 1 and this as 2. Now on adding 1 and 2, we get i plus i equals 2 into integral of log 2 cos square x pi 2 from 0 to pi by 2 plus 2 into integral of log 2 sine square x pi 2 from 0 to pi by 2. Now this implies 2 i is equal to 2 into integral of log 2 cos square x pi 2 from 0 to pi by 2 plus integral of log 2 sine square x pi 2 from 0 to pi by 2. We know that log plus log y is equal to log x pi. By using this, we get 2 into integral of log 2 cos square x pi 2 into 2 sine square x pi 2 from 0 to pi by 2. Now this is equal to 2 into integral of log 4 sine square x pi 2 cos square x pi 2 from 0 to pi by 2. Now this is equal to 2 into integral of log 2 sine x pi 2 cos x pi 2 whole square from 0 to pi by 2. This is equal to 2 into integral of log sine x whole square from 0 to pi by 2. This is equal to 2 into integral of 2 log sine x since log a to the power m is equal to m log a. So we get integral of 2 log sine x from 0 to pi by 2. So 2i is equal to 2 into integral of 2 log sine x from 0 to pi by 2. Now this implies i is equal to 2 into integral of log sine x from 0 to pi by 2 into 2. We know that integral of fx from 0 to a is equal to integral of fa minus x from 0 to a. By using this property, we get i as 2 into integral of log sine pi by 2 minus x from 0 to pi by 2. This is equal to 2 into integral of log cos x from 0 to pi by 2. Same this equation as equation number 3 and this as 4. Now on adding 3 and 4, we get i plus i equals 2 into integral of log sine x from 0 to pi by 2 plus 2 into integral of log cos x from 0 to pi by 2. This is equal to 2 into integral of log sine x cos x from 0 to pi by 2. Into integral of log sine x cos x from 0 to pi by 2. Now this implies i is equal to integral of log sine x cos x from 0 to pi by 2. Now multiply the numerator and denominator pi 2. This is equal to integral of log 2 sine x cos x pi 2 from 0 to pi by 2. This is equal to integral of log sine 2 x pi 2 from 0 to pi by 2. We know that log x pi y is equal to log x minus y. By using this, we get integral of log sine 2 x minus log 2 from 0 to pi by 2. Now let 2x is equal to t. On differentiating both sides with respect to x, we get dt by dx as 2. This implies dx is equal to 1 by 2 into dt. Now when x is equal to pi by 2, then t is equal to pi. And then x is equal to 0, then t is also equal to 0. So now we are going to substitute t in place of 2x and 1 by 2 dt in place of gx in this integral. So now we have 1 by 2 into integral of log sine t minus log 2 dt from 0 to pi. Let dt is equal to log sine t minus log 2. Now g pi minus t is equal to log sine pi minus t minus log 2. We know that sine pi minus theta is sine theta. So this is equal to log sine t minus log 2. And this is equal to dt. In differentiating property of definite integrals, it says integral of fx from 0 to 2a is equal to 2 times integral of fx from 0 to a if f of 2a minus x is equal to fx. And this is equal to 0 if f of 2a minus x is equal to minus fx. We get i as 2 by 2 into integral of log sine t minus log 2 from 0 to pi by 2. Now this is equal to integral of log sine t minus log 2 from 0 to pi by 2. In the variable t by x, we get i as integral of log sine x minus log 2 from 0 to pi by 2. From 3, we know that i is equal to 2 into integral of log sine x from 0 to pi by 2. So now we have 2 into integral of log sine x from 0 to pi by 2 equals to integral of log sine x minus log 2 from 0 to pi by 2. Now this implies integral of log sine x from 0 to pi by 2 is equal to minus log 2 into integral of 1 from 0 to pi by 2. Now we know that i is equal to 2 into integral of log sine x from 0 to pi by 2. Now here we have integral of log sine x from 0 to pi by 2. So this means i by 2 is equal to minus log 2 into integral of 1 from 0 to pi by 2. Now this implies i is equal to minus 2 log 2 into integral of 1 from 0 to pi by 2. This is equal to minus 2 x log 2 where the lower limit is 0 and upper limit is pi by 2. Substitute the value of upper limit. So we have minus 2 pi by 2 log 2 minus minus 2 into 0 into log 2. This is equal to pi log 2 at minus sine. It's a required answer is minus pi log 2. So this completes the session. I am taking care.