 Hello and welcome to the session, let us discuss the following question, it says integrate the following function. The given function is root sine 2x into cos 2x, let us now proceed on with the solution. Let i be the integral under root sine 2x into cos 2x dx. Now here we see that the derivative of sine 2x is 2 cos 2x, so put t equal to sine 2x, so dt by dx is equal to 2 cos 2x and this implies dt is equal to 2 cos 2x dx and this again implies cos 2x is equal to cos 2x into dx is equal to dt by 2. Now cos 2x into dx is dt by 2 and t is sine 2x, so substitute all these values in the integral, so the integral becomes under root t to dt by 2. It is again equal to 1 by 2 integral t to the power 1 by 2 dt and its integral is equal to 1 by 2 t to the power 1 by 2 plus 1 upon 1 by 2 plus 1 plus c. As we know that the integral of x to the power n dx is equal to x to the power n plus 1 upon n plus 1 plus c, so here n is 1 by 2 which is again equal to 1 by 2 t to the power 3 by 2 upon 3 by 2 plus c and this is again equal to 2 by 2 into 3 into t to the power 3 by 2 plus c, 2 gets cancelled with 2 and this is 1 by 3 t is sine 2x, so substitute it, it is sine 2x to the power 3 by 2 plus c. Hence the integral of the given function is 1 by 3 sine 2x to the power 3 by 2 plus 3 and this completes the question. Bye for now, take care and have a good day.