 Hello and welcome to the session. Let us discuss the following question. Let's say integrate the following function The given function is root 10x upon sine x into cos x. Let us now proceed on with the solution and let I be the integral root tan x upon sine x into cos x dx Now we know that the derivative of tan x is secant square x. So we need to have secant square x here so we multiply the numerator and the denominator by secant square x this becomes root tan x into secant square x upon sine x cos x into secant square x dx Now we know that secant x is equal to 1 upon cos x this is root tan x into secant square x upon sine x into cos x into 1 upon cos square x dx So this is further equal to under root tan x into secant square x upon the sine x upon cos x is tan x dx now we substitute t for tan x. So dt by dx is secant square x and this implies dt is equal to secant square x dx So secant square x into dx is dt so Substituting all these values in the integral. So the integral becomes under the root t upon which is again equal to t to the power minus 1 by 2 dt this t to the power 1 by 2 and this is t to the power 1. So 1 by 2 minus 1 is minus 1 by 2. Now its integral is t to the power minus 1 by 2 plus 1 upon minus 1 by 2 plus 1 plus c Which is again equal to t to the power 1 by 2 upon 1 by 2 plus c So this is equal to 2 into root t plus c. Now t is tan x. So this becomes 2 into root tan x plus c And here we have applied the formula for the integral of x to the power n dx Which is equal to x to the power n plus 1 upon n plus 1 plus c Hence the integral of the given function is 2 into root tan x plus c. This completes the question. Why for now take care and have a good day.