 Hello friends, let's solve the following problem of integration. We have to integrate the function 1 upon 1 minus tan x Let us now move on to the solution let I be the integral 1 upon 1 minus tan x d x Now we know that tan x is sine x upon cos x So this is equal to 1 upon 1 minus Sine x upon cos x dx which is further equal to cos x upon cos x minus Sine x dx taking the Lcm now We multiply and divide by 2 So we have 1 by 2 into 2 cos x upon cos x minus Sine x dx as this is written as 1 by 2 integral cos x minus sine x plus cos x plus sine x We are adding in subtracting sine x and we are writing to cos x as cos x plus cos x upon cos x minus sine x dx Now this can be further written as 1 by 2 integral cos x minus sine x upon cos x minus sine x Plus 1 by 2 integral cos x plus Sine x upon cos x minus Sine x dx Now cos x minus sine x gets cancelled with cos x minus sine x So the integral becomes 1 by 2 integral 1 dx plus 1 by 2 cos x plus Sine x upon cos x minus Sine x dx Now here we see that the derivative of cos x minus sine x is minus of cos x plus sine x so put equal to cos x minus Sine x so dt by dx is equal to minus sine x minus cos x So this implies dt is equal to minus of cos x plus sine x dx and this implies cos x plus sine x dx is equal to minus dt So cos x plus sine x into dx is equal to minus dt and Cos x minus sine x is t so substituting all these values in the second integral we have integral as 1 by 2 dx plus 1 by 2 integral minus dt upon t Which is again equal to 1 by 2 integral dx minus 1 by 2 integral 1 by t dt Now the value of the first integral is 1 by 2 x minus 1 by 2 integral 1 by t dt is log mod t plus c where c is the constant of integral Now t is cos x minus sine x so let us substitute integral of the given function is x by 2 minus 1 by 2 log mod cos x minus sine x plus c and this completes the question and the session Keep practicing such questions Try to use the substitution method So by for now take care. Have a good day