 Hello and welcome to the session. Let us discuss the following question. It says integrate the following function. The given function is 2 cos x minus 3 sin x upon 6 cos x plus 4 sin x. Let us now proceed on with the solution. And let i be the integral 2 cos x minus 3 sin x upon 6 cos x plus 4 sin x dx. Again, this is equal to integral 2 cos x minus 3 sin x Taking to common from the denominator it becomes 2 into 3 cos x plus 2 sin x dx. Now we see that the derivative of 3 cos x plus 2 sin x is 2 cos x minus 3 sin x. So put equal to 3 cos x plus 2 sin x. So dt is equal to minus 3 sin x plus 2 cos x because derivative of cos x is minus sin x and derivative of sin x is cos x into dx. That is, dt is equal to 2 cos x minus 3 sin x dx. So 2 cos x minus 3 sin x dx is equal to dt and 3 cos x plus 2 sin x is t. Substituting all these values in the integral, the integral i becomes dt upon 2 which can be again written as 1 by 2 integral of 1 by t dt. Now the integral of 1 by t dt is log mod t plus t. So this is 1 by 2 log mod t plus c. Let's now substitute the value of t. This is 1 by log mod 2 sin x plus 3 cos x plus c. Hence integral of the given function is 1 by 2 log mod 2 sin x plus 3 cos x plus c. And this completes the question. Hope you enjoy this session. Goodbye and take care.