 Hi, and welcome to our session. Let us discuss the following question. The question says, by using the properties of definite integrals, evaluate the following integral of 2 log sine x minus log sine 2x from 0 to pi by 2. Before solving this question, we should know that integral of fx from 0 to a is equal to integral of f a minus x from 0 to a. The knowledge of this property is a key idea in this question. With the solution, let i is equal to integral of 2 log sine x minus log sine 2x from 0 to pi by 2. We know that sine 2x is equal to 2 sine x into cos x. This implies i is equal to integral of 2 log sine x minus log 2 sine x cos x from 0 to pi by 2. We know that log xyz is equal to log x plus log y plus log z. So using this result, we get i as integral of 2 log sine x minus log 2 minus log sine x minus log cos x from 0 to pi by 2. This implies i is equal to integral of 2 log sine x minus log sine x minus log 2 minus log cos x from 0 to pi by 2. Now by using property of definite integrals which says integral of fx from 0 to a is equal to integral of f minus x from 0 to a. We get i as integral of log. Now here a is equal to pi by 2. And fx is equal to log sine x minus log 2 minus log cos x. So f of a minus x is log sine pi by 2 minus x minus log 2 minus log cos pi by 2 minus x from 0 to pi by 2. We know that sine pi by 2 minus x is cos x and cos pi by 2 minus x is sine x. So using this, we have integral of log cos x minus log 2 minus log sine x from 0 to pi by 2. Let's name this equation as equation number 1 and this as 2. On adding 1 and 2, we get plus i equals to integral of log sine x minus log 2 minus log cos x from 0 to pi by 2 plus integral of log cos x minus log 2 minus log sine x from 0 to pi by 2. And this is equal to integral of minus 2 log 2 from 0 to pi by 2. This is equal to minus 2 log 2 into integral of 1 from 0 to pi by 2. Now this implies 2i is equal to minus 2 log 2 into x, where the lower limit is 0 and upper limit is pi by 2. So we will use second fundamental theorem of integral calculus. So we will first substitute the upper limit. Upper limit is pi by 2. By substituting pi by 2, we get minus 2 log 2 into pi by 2 minus minus 2 log 2 into 0. This is equal to minus 2 into pi by 2 into log 2. And this is equal to minus pi log 2. So 2i is equal to minus pi log 2. This implies i is equal to minus pi by 2 log 2. Now this can be written as pi by 2 inverse. And this can be written as pi by 2 log 1 by 2. Hence our required answer is pi by 2 log 1 by 2. So this completes the session. Bye and take care.