 Hi and welcome to the session, let's work out the following question. The question says evaluate integral tan inverse square root 1 minus sin x divided by 1 plus sin x dx. So let's start with the solution to this question, let y be equal to integral tan inverse square root 1 minus sin x divided by 1 plus sin x dx. This is equal to integral tan inverse square root 1 minus cos pi by 2 minus x divided by 1 plus cos pi by 2 minus x. Now we can write this because we see that sin x is same as cos pi by 2 minus x. This can be further written as integral tan inverse square root. Now the numerator can be written as 2 sin square pi by 4 minus x by 2 divided by the denominator that can be written as 2 cos square pi by 4 minus x by 2 into dx that is equal to integral tan inverse to tan pi by 4 minus x by 2 dx because 2 gets cancelled with 2 sin square pi by 4 minus x by 2 divided by cos square pi by 4 minus x by 2 is same as tan square pi by 4 minus x by 2 and when that comes out of the square root it becomes tan pi by 4 minus x by 2. This implies that y is equal to integral pi by 4 minus x by 2 dx that is equal to pi by 4 x minus x square by 4 plus c. So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.