 Hi and welcome to our session. I am Kanika and I am going to help you to solve the following question. The question says by using the properties of definite integrals evaluate the following. Integral of x by 1 plus sin x from 0 to pi. Before solving this question we should know that integral of f x from 0 to a is equal to integral of f a minus x from 0 to a. The knowledge of this property is the key idea in this question. Looking with the solution let i is equal to integral of x by 1 plus sin x from 0 to pi. From the key idea we know that integral of f x from 0 to a is equal to integral of f a minus x from 0 to a. By using this property we get i as now here is equal to pi i is equal to integral of pi minus x by 1 plus sin pi minus x where the lower limit is 0 and upper limit is pi. We know that sin pi minus x is equal to sin x equals to integral of pi minus x by 1 plus sin pi minus x from 0 to pi implies i is equal to integral of pi minus x upon 1 plus sin x from 0 to pi. Let's name this equation as equation number 1 and this as 2. Now on adding 1 and 2 we get plus i equals to integral of x by 1 plus sin x from 0 to pi plus integral of pi minus x by 1 plus sin x from 0 to pi. Now this is equal to integral of x plus pi minus x by 1 plus sin x from 0 to pi and this is equal to integral of pi by 1 plus sin x from 0 to pi. Now we will multiply the numerator and denominator by 1 minus sin x. So we have integral of pi by 1 plus sin x into 1 minus sin x by 1 minus sin x where the lower limit is 0 and upper limit is pi. Now this is equal to pi into integral of 1 minus sin x by 1 minus sin square x from 0 to pi. We know that 1 minus sin square x is equal to cos square x this is equal to pi into integral of 1 minus sin x by cos square x from 0 to pi. This is equal to pi into integral of 1 by cos square x minus sin x by cos square x from 0 to pi. We know that 1 by cos theta is secant theta so this is equal to pi into integral of secant square x minus. Now sin x by cos square x can be written as sin x by cos x into 1 by cos x sin x by cos x is tan x and 1 by cos x is secant x. So we have secant square x minus secant x tan x lower limit is 0 and upper limit is pi. Now this is equal to pi into integral of secant square x from 0 to pi minus pi into integral of secant x tan x from 0 to pi. We know that integral of secant square x with respect to x is tan x and integral of secant x tan x with respect to x is secant x. So by using these formulas we get pi into tan x lower limit is 0 and upper limit is pi minus pi into secant x lower limit is 0 and upper limit is pi. Now by using second fundamental theorem of integral calculers we will first calculate value of tan x at pi. So we have pi into tan pi minus. Now we will calculate value of tan x at 0 so we have tan 0 minus pi into secant pi minus secant 0. Now tan pi is equal to 0 and tan 0 is also 0 minus pi into secant pi is minus 1 and secant 0 is 1. So we have minus pi into minus 2 and this is equal to 2 pi. So i plus i is equal to 2 pi this implies 2i is equal to 2 pi and this implies i is equal to pi. Hence our required answer is pi. So this completes the session. Bye and take care.