Hi and welcome to the session. Let us discuss the following question. Questions is integrate the following functions. Given function is sex square x upon under root of tan square x plus 4. First of all, let us understand that integral dx upon square root of x square plus a square is equal to log of x plus square root of x square plus a square plus c, where c is the constant of integration. Now we will use this formula as our key idea to solve the given question. Let us now start with the solution. Now we have to find integral of sex square x dx upon square root of tan square x plus 4. Now in the integrand, clearly we can see derivative of tan x is equal to sex square x. So we will substitute tan x is equal to t. Now differentiating both the sides with respect to x, we get sex square x dx is equal to dt. So we get integral sex square x dx upon square root of tan square x plus 4 is equal to integral of dt upon square root of t square plus 4. Now 4 can be written as 2 square. So this integral is equal to integral of dt upon t square plus 2 square. Now using the formula given in key idea, we get this integral is equal to log of t plus square root of t square plus 2 square plus c, where c is the constant of integration. Comparing this integral with the integral given in the key idea, clearly we can see variable x has been replaced by variable t and a is equal to 2. So this integral is equal to log of t plus square root of t square plus 2 square plus c. Now simplifying further, we can write it as log of t plus square root of t square plus 4 plus c. Now we know t is equal to tan x. So we will substitute tan x for t in this expression and we get log of tan x plus square root of tan square x plus 4 plus c. So our required answer is integral of sec square x dx upon square root of tan square x plus 4 is equal to log of tan x plus square root of tan square x plus 4 plus c. This completes the session. Hope you understood the solution. Take care and have a nice day.