 Hello, and how are you all? Question says integrate the following. Now the function which is given to us is 1 divided by sin x cos cube x. Now let us integrate this function with respect to dx. Now, we know that 1 can be written as sin square x plus cos square x, right? So in substituting it we have sin square x plus cos square x divided by sin x cos cube x into dx. Now here we can write it also as sin square x divided by sin x cos cube x plus into dx cos square x divided by sin x cos cube x into dx. In simplifying we have the answer as sin x divided by cos cube x dx plus 1 over sin x cos x dx. Now here we can write this function as tan x into we are left with 1 upon cos square x which can be written as sin square x into dx plus here we can write sin square x divided by tan x. Now let t be equal to tan x. So dt will be equal to sin square x into dx, right? So on substituting we have t into dt plus dt over that is equal to here integral of t into dt is t square divided by 2 plus integral of dt by t is log mod of t plus c. Now on substituting the value of t as tan x we have tan square x divided by 2 plus log of mod tan x plus c which can be written also as on rearranging log of mod of tan x plus 1 by 2 tan square x plus so these are the required answer to the question. Hope you understood it well and enjoy it too. Have a nice day.