 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 sin square x upon sin x plus cos x dx. Let us see how we can evaluate this integral. First of all we assume let i be equal to the given integral which is integral 0 to pi by 2 sin square x upon sin x plus cos x dx. We have a property which says that integral 0 to a fx dx is equal to integral 0 to a f of a minus x dx. So using this property in the above given i we get i is equal to integral 0 to pi by 2 sin square so in place of x we will put a minus x that is pi by 2 minus x. So in the numerator it would be sin square pi by 2 minus x upon sin pi by 2 minus x plus cos pi by 2 minus x dx. Now this becomes i is equal to integral 0 to pi by 2 cos square x upon cos x plus sin x dx. Since we know that sin pi by 2 minus x is equal to cos x and cos pi by 2 minus x is equal to sin x and so sin square pi by 2 minus x is equal to cos square x. So using these results we get i is equal to integral 0 to pi by 2 cos square x upon cos x plus sin x dx. Now we assume this as equation 1 and this as equation 2. Now next we shall add equations 1 and 2. So adding equations 1 and 2 we get i plus i that is 2i is equal to integral 0 to pi by 2 sin square x upon sin x plus cos x dx plus integral 0 to pi by 2 cos square x upon cos x plus sin x dx. So this becomes i is equal to integral 0 to pi by 2 sin square x plus cos x dx. Now we know that sin square x plus cos square x is equal to 1 therefore using this result in this 2i we get 2i is equal to integral 0 to pi by 2 dx upon sin x plus cos x since we have the sin square x plus cos square x is equal to 1. Now let's write sin x and cos x in terms of tan x upon 2. So sin x is equal to 2 tan x upon 2 upon 1 plus tan square x upon 2 and cos x can be written as 1 minus tan square x upon 2 upon 1 plus tan square x upon 2. Using these values of sin x we get 2i is equal to integral 0 to pi by 2 dx upon. Now in place of sin x we will use this value that is x upon 2 upon 1 plus tan square x upon 2 plus cos x. We will use this value for cos x that is 1 minus tan square x upon 2 upon 1 plus tan square x upon. Now we take let upon 2 be equal to t. Now differentiating both sides with respect to x we get 1 upon 2 into sin square x upon 2 dx is equal to dt which means that dx is equal to 2 dt upon sin square x upon 2 which is 1 plus tan square x upon 2 that is 1 plus t square since we are taken tan x upon 2 as t and we know that 1 plus tan square x is equal to 6 square x. So we have got the value for dx. Now the limits will also change accordingly like when we have x is equal to 0 then t would be equal to tan 0 since t is equal to tan x upon 2 and in place of x we put 0 and this is equal to 0. So when x is equal to 0 t is equal to 0. Now when x is equal to the upper limit that is pi by 2 then t would be equal to tan pi by 4 since t is equal to tan x upon 2 and in place of x we put pi by 2 and tan pi by 4 is 1. So we get t is equal to 1 then x is equal to pi by 2. So now the limits will change from 0 to pi by 2 to 0 to 1 thus we get 2i is equal to integral 0 to 1. Now in place of dx we would put 2 dt upon 1 plus t square so 2 upon 1 plus t square dt and this will upon 2t upon 1 plus t square that is in place of tan x upon 2 here we will put t plus 1 minus t square upon 1 plus t square. So this gives us 2i is equal to integral 0 to 1 2dt upon 2t plus 1 minus t square. This means we have 2i is equal to minus 2 integral 0 to 1 dt upon t square minus 2t minus 1 that is we take minus common from the denominator and 2 common from the numerator. So outside the integral we get minus 2. So this gives us 2i is equal to minus 2 integral 0 to 1 dt upon t square minus 2t the minus 1 can be written as minus 2 plus 1 so this further gives us 2i is equal to minus 2 integral 0 to 1 dt upon now t square minus 2t plus 1 is t minus 1 whole square minus 2 can be written as root 2 whole square. Now we can cancel this 2 with this 2 so we get i is equal to minus integral 0 to 1 dt upon t minus 1 whole square minus root 2 whole square. We have the formula integral dx upon x square minus a square is equal to 1 upon 2a log modulus x minus a upon x plus a. So using this formula in this value for i we get i is equal to minus 1 upon 2 into a now here a is root 2 log modulus x minus a now x in this case would be t minus 1 so t minus 1 minus root 2 upon x plus a that is t minus 1 plus root 2 and here the limits are from 0 to 1 this means we get i is equal to minus 1 upon 2 root 2 log modulus. Now first we will put the upper limit that is 1 1 minus 1 minus root 2 upon 1 minus 1 plus root 2 minus log modulus now we will put the lower limit that is 0 in place of t 0 minus 1 minus root 2 upon 0 minus 1 plus root 2 from here we have i is equal to minus 1 upon 2 root 2 log modulus minus root 2 upon root 2 minus log modulus 1 plus root 2 upon 1 minus root 2 this further gives us i equal to minus 1 upon 2 root 2 log modulus minus 1 minus log modulus 1 plus root 2 upon 1 minus root 2 this further gives us i equal to minus 1 upon 2 root 2 this whole into 0 minus log modulus 1 plus root 2 upon 1 minus root 2 now since we know that log of modulus minus 1 is equal to log 1 which is equal to 0 so here we have got 0 now from here we have i is equal to minus 1 upon 2 root 2 into minus log minus modulus 1 plus root 2 upon 1 minus root 2 now we multiply the numerator and denominator by 1 plus root 2 now from here we have i is equal to 1 upon 2 root 2 into log modulus 1 plus root 2 whole square upon minus 1 so this gives us i is equal to 1 plus root 1 upon modulus 1 plus root 2 whole square again since we know that log of modulus minus 1 is equal to log 1 which is equal to 0 from here we get i is equal to 2 upon 2 root 2 into log 1 plus root 2 since we know log a to the power n is equal to n into log a now this 2 cancels with this 2 so we get i is equal to 1 upon root 2 log 1 plus root 2 this is the value for and as we get integral 0 to pi by 2 sin square x is equal to 1 upon root 2 log 1 plus so with this we complete this session hope you have understood the solution for this question