 Hello friends, let's work out the following problem. It says integrate the following function. The given function is cos root x upon root x. Let us now proceed on with the solution and let i be the integral cos root x upon root x dx. Now here we see that the derivative of root x is 1 by 2 root x which contains 1 by root x that is this thing. So put equal to root x. So dt by dx is equal to 1 by 2 root x dx because it can be written as x to the power 1 by 2. So the derivative is 1 by 2 into x to the power 1 by 2 minus 1 which is x to the power minus 1 by 2. So this is equal to 1 by 2 root x. And this implies dt is equal to 1 by 2 root x dx. So 1 by root x is 2 dt. 2 dt is equal to 1 by root x dx. Now 1 by root x dx is 2 dt and root x is t. So substituting all these values in the integral becomes cos t into 2 dt. So the integral becomes 2 cos t dt. Now we know that the integral of cos t is sin t plus c. So this is equal to 2 sin t plus c. Let's now substitute the value of t. So this becomes 2 sin root x plus c. Hence the integral of the given function is 2 sin root x plus c. And this completes the question. Bye for now. Take care. Have a good day.