 Hello and welcome to the session. Let's solve the following question. It says integrate the following function The given function is cot x into log sine x. Let us now proceed on with the solution and let I be the integral cot x into log sine x dx Now here we see that the derivative of log sine x is cot x. So put y equal to log sine x So dy by dx is equal to 1 upon sine x derivative of log sine x is 1 upon sine x into derivative of sine x and we know that the derivative of sine x is cos x. So this becomes cos x upon sine x which is equal to cot x and this implies dy is equal to cot x dx. So cot x into dx is dy and y is log sine x. So substitute all these values in the integral. The integral becomes y dy and the value of the integral is y square upon 2 plus c. As we know that the integral of y to the power n dy is equal to y to the power n plus 1 upon n plus 1 plus c. Now here n is 1 So the integral is y to the power 2 upon 2 plus c. Now let's substitute the value of y here. So the integral has value log sine x square plus c. Hence the integral of the given function is 1 by 2 log sine x whole square plus c. And this completes the question. Bye for now. Take care. Have a good day.