 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 sin 2x dx. Let us first understand second fundamental theorem of integral calculus. Two levels are to evaluate definite integral by making use of antiderivative. Let a continuous function define on the closed interval a b and f derivative of f. Then rule of fx dx is equal to the derivative of f and this is equal to minus that is the integral from a to b fx dx is equal to the value of the antiderivative at unit b minus the value of the same antiderivative at the lower limit a. This is the key idea behind that question. We will take the help of this key idea to solve the above question. So let's start the solution is equal to a given integral 0 to pi by 4 sin 2x dx. Find that in definite integral sin 2x. Now we know that is equal to 2 sin theta. This in definite integral can be written as sin x cos x in definite integral is equal to sin x cos x dx is equal to is equal to and this is equal to t square. Now we will re substitute the value of t is equal to sin square x. It is a antiderivative a given function that is sin 2x fundamental theorem is equal to is equal to pi by 4 sin square 0. Now sin pi by 4 is equal to 1 over root is equal to 1 over root 2 square minus this is equal to 1 over 2. For the other question is 1 over 2. I hope the solution is clear to you. Bye and take care.