 Hello and welcome to the session, the given question says evaluate the following definite integral, ninth is integral cos 2x dx and the lower limit of integration is 0 and upper limit is pi by 2. So let's start with the solution and first let us find the value of integral cos 2x dx. Let t is equal to 2x, so dt is equal to 2 into dx or dx is equal to dt upon 2. So this can further be written as integral cos t into dt upon 2 or we have half integral cos t into dt. This is further equal to half sin t. Now substituting the value of t that is 2x we have half sin 2x. Now by the second fundamental theorem of integral calculus we have the given definite integral can be written as sin 2x the lower limit is 0 and upper limit is pi by 2. So this is further equal to taking half common first putting the upper limit we have sin 2 into pi by 2 minus sin 2 into 0 which is further equal to half. Cancelling we have sin pi and its value is 0 minus sin 0 is again 0. So we have half into 0 which is equal to 0. That's on integrating the given definite integral we get its value as 0. So this completes the solution by integral.