 Hello and welcome to the session. In this session, we discussed the following question which says evaluate integral 0 to pi by 2 cos x upon 3 cos x plus sin x dx. Let's see how we evaluate this integral. First of all, we assume that I be equal to integral 0 to pi by 2 cos x upon 3 cos x plus sin x dx. Now we can write this numerator that is cos x equal to a into the denominator that is 3 cos x plus sin x where this a is some constant plus some constant b into b by dx of the denominator which is 3 cos x plus sin x. So this means we get cos x is equal to 3 a into cos x plus a into sin x plus b into Now d by dx of 3 cos x would be minus 3 sin x plus d by dx of sin x would be cos x. So we get cos x is equal to 3 a plus b into cos x plus a minus 3 b into sin x. Now comparing the coefficients of we get 3 a plus b is equal to 1. Let this be equation 1 then comparing the coefficients of sin x we get a minus 3 b is equal to 0. Let this be equation 2. So we get 2 equations 3 a plus b equal to 1 and a minus 3 b equal to 0. This is equation 1. This is equation 2. We solve both these equations to get the values for a and b. Now from equation 2 we have a is equal to 3 b. So now substituting a equal to 3 b in equation 1 we get 3 into 3 b plus b is equal to 1 that is 9 b plus b equal to 1 which means 10 b is equal to 1 or we get b is equal to 1 upon 10. Now substituting b equal to 1 upon 10 in a equal to 3 b we get a is equal to 3 into 1 upon 10 that is equal to 3 upon 10. So we can rewrite cos x is equal to a into the denominator which was 3 into cos x plus sin x plus b into d by dx of the denominator which was minus 3 sin x plus cos x. Now putting the values of a and b we get cos x is equal to 3 upon 10 into 3 cos x plus sin x plus b which is 1 upon 10 into minus 3 sin x plus cos x. And we have taken this i is equal to integral 0 to pi by 2 cos x dx upon 3 cos x plus sin x. So we can now write i is equal to integral 0 to pi by 2 3 upon 10 into 3 cos x plus sin x upon 3 cos x plus sin x dx plus integral 0 to pi by 2 1 upon 10 minus 3 sin x plus cos x upon 3 cos x plus sin x dx. Now here 3 cos x plus sin x cancels so this is equal to integral 0 to pi by 2 3 upon 10 dx plus 1 upon 10 integral 0 to pi by 2 minus 3 sin x plus cos x upon 3 cos x plus sin x dx. So this is equal to 3 upon 10 into x and the limit goes from 0 to pi by 2 plus 1 upon 10 integral 0 to pi by 2 minus 3 sin x plus cos x upon 3 cos x plus sin x dx. This is our i. Now to solve the second integral we would take 3 cos x plus sin x equal to t differentiating both sides with respect to x we get minus 3 sin x plus cos x dx is equal to dt. Therefore we get integral 0 to pi by 2 minus 3 sin x plus cos x upon 3 cos x plus sin x dx is equal to now the limits will change accordingly. So then we take x equal to 0 then t is equal to 3 cos 0 plus sin 0 that is equal to 3 and when we take x equal to pi by 2 then t would be equal to 3 cos pi by 2 plus sin pi by 2 that is equal to 3 into 0 plus 1 equal to 1. So this integral is equal to integral 3 to 1 minus 3 sin x plus cos x dx is dt upon t which is 3 cos x plus sin x. This is equal to minus integral 1 to 3 dt upon t which is equal to minus log modulus t limit goes from 1 to 3 that is equal to minus log modulus 3 minus log modulus 1 which is equal to minus log 3 that is integral 0 to pi by 2 minus 3 sin x plus cos x upon 3 cos x plus sin x dx is equal to minus log 3. We would put the value of this integral as minus log 3 in i that is here so we would get i is equal to 3 upon 10 into x limit goes from 0 to pi by 2 minus 1 upon 10 into log 3 which is the value of this integral. So i is equal to 3 upon 10 into pi by 2 minus 0 minus 1 upon 10 log 3 so we have i is equal to 3 pi by 20 minus log 3 upon 10. This is the value of i or you say integral 0 to pi by 2 cos x upon 3 cos x plus sin x dx is equal to 3 pi by 20 minus log 3 upon 10. So this is our final answer. This completes the session. Hope you have understood the solution for this question.