 Hello and welcome to the session. In this session we discussed the following question which says evaluate the following integral 0 to pi by 2 dx upon 2 cos x plus 4 sin x. Let us try to evaluate this integral. We take let i be equal to integral 0 to pi by 2 dx upon 2 cos x plus 4 sin x. Now we know that cos 2 x is equal to cos square x minus sin square x. So this means we can write cos x equal to cos square x upon 2 minus sin square x upon 2. Also we know that sin 2 x is equal to 2 sin x into cos x. So this would mean that sin x is equal to 2 sin x upon 2 into cos x upon 2. So now we shall substitute these values of cos x and sin x in this i. So we get i is equal to integral 0 to pi by 2 dx upon 2 into cos x which is cos square x upon 2 minus sin square x upon 2 plus 4 into sin x which is 2 into sin x upon 2 into cos x upon 2. This means we get i is equal to integral 0 to pi by 2 dx upon 2 cos square x upon 2 minus 2 sin square x upon 2 plus 8 sin x upon 2 into cos x upon 2. Next we will divide the numerator and denominator by cos square x upon 2. So dividing numerator and denominator by cos square x upon 2 we get i is equal to integral 0 to pi by 2. So x square x upon 2 dx upon 2 minus 2 square x upon 2 plus 8 into tan x upon 2. Now since we know that 1 upon cos square x upon 2 is 6 square x upon 2. So when we divide the numerator by cos square x upon 2 we get 6 square x upon 2 and in the denominator we divide each term by cos square x upon 2. So when 2 cos square x upon 2 is divided by cos square x upon 2 we get 2. Then consider the next term that is 2 sin square x upon 2. This when divided by cos square x upon 2 gives us 2 tan square x upon 2 plus 8 into sin x upon 2 into cos x upon 2 upon cos square x upon 2 gives us 8 sin x upon 2 upon cos x upon 2 that is equal to 8 tan x upon 2. Next putting tan x upon 2 as t. Now differentiating both the sides with respect to x we get 1 upon 2 into 6 square x upon 2 dx is equal to dt that is 6 square x upon 2 dx is equal to 2 dt. Therefore we get i is equal to integral. Now when we take x equal to 0 then t is equal to tan 0 that is equal to 0 and when we take x equal to pi by 2 then we get t is equal to tan pi by 4 which is equal to 1. So now the limits changes from 0 to pi by 2 to 0 to 1 thus i is equal to integral 0 to 1. In the numerator we would have 2 dt since we had 6 square x upon 2 into dx which is equal to 2 dt so 2 dt upon 2 minus 2 t square plus 8 t that is instead of tan x upon 2 we put t this gives us i is equal to integral 0 to 1 dt upon minus t square minus 40 minus 1. Now if we cancel 2 from the numerator and the denominator. So this means we get i is equal to minus integral 0 to 1 dt upon t square minus 40 minus 1 this further gives us i is equal to minus integral 0 to 1 dt upon t square minus 40 and we can write minus 1 as plus 4 minus 5 so this further gives us i equal to minus integral 0 to 1 dt upon t square minus 40 plus 4 minus 5 further we have i equal to minus integral 0 to 1 dt upon now we can factorize t square minus 40 plus 4 by splitting the middle term so this is equal to t square minus 2 t minus 2 t plus 4 that is we take t common inside the bracket we have t minus 2 and we take minus 2 common inside the bracket we have t minus 2 so this is equal to t minus 2 into t minus 2 that is equal to t minus 2 to whole square. So t square minus 4t plus 4 can be written as t minus 2 whole square minus 5 which can be written as root 5 whole square. We have a formula integral dx upon x square minus a square is equal to 1 upon 2a into log modulus x minus a upon x plus a plus c. Therefore, using this formula in the above i that we have obtained, we have i is equal to minus 1 upon 2a. Now here a is root 5 so 2 into root 5 into log modulus x, x would be t minus 2 minus a that is root 5 upon x plus a that is t minus 2 plus root 5 and the limits are from 0 to 1. So i is equal to minus 1 upon 2 root 5 this into log. Now putting t as 1 we get 1 minus 2 minus root 5 upon 1 minus 2 plus root 5 minus log. Now we put t as 0, 0 minus 2 minus root 5 upon 0 minus 2 plus root 5. So this gives us i is equal to minus 1 upon 2 root 5 into log modulus minus 1 minus root 5 upon minus 1 plus root 5 minus log modulus minus 2 minus root 5 upon minus 2 plus root 5. This gives us i is equal to minus 1 upon 2 root 5 into log 1 plus root 5 upon root 5 minus 1 minus log 2 plus root 5 upon root 5 minus 2. Now this can be further written as i equal to 1 upon 2 root 5 into log 2 plus root 5 upon root 5 minus 2 minus log 1 plus root 5 upon root 5 minus 1. This further gives us i is equal to 1 upon 2 root 5 into log 2 plus root 5 upon root 5 minus 2 multiplied by root 5 minus 1 upon 1 plus root 5. Since we know log of a minus log of b is equal to log of a upon b. So log 2 plus root 5 upon root 5 minus 2 minus log 1 plus root 5 upon root 5 minus 1 is given by log of 2 plus root 5 upon root 5 minus 2 divided by this or multiplied by the reciprocal of this. This gives us i is equal to 1 upon 2 root 5 into log 2 plus root 5 when multiplied by root 5 minus 1 gives us 3 plus root 5 upon 1 plus root 5 multiplied by root 5 minus 2 which gives us 3 minus root 5. So this gives us i equal to 1 upon 2 root 5 into log 3 plus root 5 whole square upon 3 minus root 5 into 3 plus root 5. That is here we multiply the numerator and denominator by 3 plus root 5 from here we get i equal to 1 upon 2 root 5 log 3 plus root 5 whole square upon 4. That is 3 minus root 5 whole multiplied by 3 plus root 5 is equal to 9 plus 3 root 5 minus 3 root 5 minus 5 which is equal to 9 minus 5 4. So we have got this 4 here. This gives us i is equal to 1 upon 2 root 5 into log 3 plus root 5 upon 2 whole square and from here we get i is equal to 2 upon 2 root 5 log 3 plus root 5 upon 2. Since we know log of a to the power n is equal to n into log a. So from here we get this 2 and this 2 cancels. So i is equal to 1 upon root 5 into log 3 plus root 5 upon 2. So the value of the integral 0 to pi by 2 dx upon 2 cos x plus 4 sin x is equal to 1 upon root 5 into log 3 plus root 5 upon 2. So with this we complete this session hope you have understood the solution for this question.