 Hello students, let's solve the following problem on integration. We have to integrate the function 1 upon cos square x into 1 minus tan x whole square. Let us now move on to the solution. And let i be the integral 1 upon cos square x into 1 minus tan x whole square. Now we know that 1 upon cos x is secant x. So 1 upon cos square x is secant square x. So this integral can be written as secant square x upon 1 minus tan x whole square dx. Here also we need to write dx. Now we see that the derivative of 1 minus tan x is minus secant square x. So put equal to 1 minus tan x. So dt by dx is equal to minus secant square x and this implies dt is equal to minus secant square x dx. And this implies secant square x dx is equal to minus dt. Now secant square x dx is minus dt and 1 minus tan x is t. So substitute all these values in the integral. The integral i becomes minus dt upon t to the power 2 which is again equal to minus integral t to the power minus 2 dt. Now this integral is equal to minus t to the power minus 2 plus 1 upon minus 2 plus 1 plus c which is equal to minus t to the power minus 1 upon minus 1 plus c which is equal to 1 upon t plus c. Now substitute the value of t. t is 1 minus tan x and here we have used the formula that integral of x to the power n dx is x to the power n plus 1 upon n plus 1 plus c. Now here n is minus 2. Hence the integral of the given function is 1 upon 1 minus tan x plus c. So this completes the question and the session. Bye for now. Take care. Have a good day.