 Hi and welcome to the session. Let us discuss the following question. Question says, for each of the differential equations, find the general solution. Given differential equation is, dy upon dx is equal to 1 minus cos x upon 1 plus cos x. Let us now start with the solution. Now we have to find the general solution of differential equation, dy upon dx is equal to 1 minus cos x upon 1 plus cos x. Now we know 1 minus cos x is equal to 2 sin square x upon 2 and 1 plus cos x is equal to 2 cos square x upon 2. Now 2 and 2 will get cancelled and we get dy upon dx is equal to sin square x upon 2 upon cos square x upon 2. Now we know sin x upon cos x is equal to tan x. So sin x upon 2 upon cos x upon 2 is equal to tan x upon 2 and we get dy upon dx is equal to tan square x upon 2. Now let us name this equation as 1. Now separating the variables in equation 1 we get dy is equal to tan square x upon 2 dx. Now integrating both the sides we get integral of dy is equal to integral of tan square x upon 2 dx. Now we know tan square theta is equal to sin square theta minus 1. So tan square x upon 2 can be written as sin square x upon 2 minus 1. So now we can write integral of dy is equal to integral of sin square x upon 2 minus 1 dx. Substituting sin square x upon 2 minus 1 for tan square x upon 2 in this integral we get this integral. Now this integral can be further written as integral of sin square x upon 2 dx minus integral of dx. Now we will evaluate all these integrals. Now we know integral of dy is equal to y. Now we will find integral of sin square x upon 2 dx. We can find integral of sin square x upon 2 dx by substitution method. Now put x upon 2 equal to t. This implies x is equal to 2t. Now differentiating both the sides with respect to x we get 1 is equal to 2 dt upon dx. Now this implies dx is equal to 2dt. Now this integral is equal to 2 multiplied by integral of sin square t dt. Now we know integral of sin square x is equal to tan x. So integral of sin square t with respect to t is equal to tan t. So here we can write this integral is equal to 2 tan t plus c. Now we know t is equal to x upon 2. Now we will substitute x upon 2 for t here and we get 2 tan x upon 2 plus c is equal to integral of sin square x upon 2 dx. Now we can write integral of sin square x upon 2 dx is equal to tan x upon 2 multiplied by 2 minus. Now integral of 1 with respect to x is equal to x only. So here we can write x plus c where c is the constant of integration. So this is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.