 Hello and welcome to the session. I am Deepika here. Let's discuss the question which says, evaluate the following definite integral integral from 0 to pi by 4 tan x dx. Let us first understand the second fundamental theorem integral calculus. Enable first to evaluate definite integral by making use of antiderivative. Let continuous function define on the closed interval in the derivative of f. f integral from a to b x dx is equal to derivative of the given function fx from a to b which is again equal to f of the value of the antiderivative f value of the same antiderivative at the lower limit a. So this is the key idea behind that. We will take the help of this key idea to solve the above question. So let's solve the solution is equal to, now we will first find the indefinite integral tan x dx. Now we can rewrite this as upon cos x t. Now here cos x is equal to t then x is equal to dt dx is equal to minus dt. The integral of tan x upon cos x dx is equal to minus log we will re-substitute the value of t that is it is equal to log of cos x and this is again equal to log of log sec x fx because it is an antiderivative of by second fundamental theorem we get t is equal to, this is equal to log of pi by 4 log of log log 1 log log m minus log n is equal to log m over n. So this is equal to log of and this is again equal to log of root 2 and we can rewrite this as log of 2 raise to power 1 by 2 and this is equal to 1 by 2 log 2 because log m raise to power n is equal to n log m. So the answer for the question is 1 by 2 log 2. I hope the solution is clear to you. Bye and take care.