 Hi friends, I am Purva and today we will work out the following question. Evaluate integral of sin x upon 1 minus cos x into 2 minus cos x dx. Let us begin with the solution now. Now let us denote integral of sin x upon 1 minus cos x into 2 minus cos x dx by i. So we have let i is equal to integral of sin x upon 1 minus cos x into 2 minus cos x dx. Now let us put minus cos x equal to t. So differentiating we get sin x dx because differentiating cos x we get minus sin x. So differentiating minus cos x we get sin x dx. This is equal to differentiating t we get dt. Now putting these values in 1 we get therefore i is equal to integral of dt upon 1 plus t into 2 plus t. Putting minus cos x equal to t in denominator here we get 1 plus t into 2 plus t in denominator and putting sin x dx is equal to dt in numerator we get dt in numerator here. Now we can also write this as this is equal to integral of dt upon 1 plus t minus integral of dt upon 2 plus t. This is further equal to now integral of dt upon 1 plus t is equal to log of mod 1 plus t minus integral of dt upon 2 plus t is equal to log of mod 2 plus t plus c where c is the constant of integration. This is since we know that integral of 1 upon x dx is equal to log of mod x. So applying this formula here we get integral of 1 upon 1 plus t dt is equal to log of mod 1 plus t and integral of 1 upon 2 plus t dt is equal to log of mod 2 plus t. Now putting this value of t here we get this is equal to log of mod 1 minus cos x minus log of mod 2 minus cos x plus c and this is since t is equal to minus cos x. Now this is further equal to log of mod 1 minus cos x upon 2 minus cos x plus c. This is since we know that log of m minus log of n is equal to log of m upon n. So we have got our answer as log of mod 1 minus cos x upon 2 minus cos x plus c. Hope you have understood the solution. Bye and take care.