So, log(sinx + cosx) = log[â2.sin(x+â/4)]
= logâ2 + logsin(x+â/4)
= (logâ2)x + I'
In I', put x+â/4 = t. So, dx = dt
Now, the limit is 0 to â/2
So, I = â/2.logâ2 + I', where the limits are from o to â/2 ... (A)
Now, I'=â« log(sin x).dx ...(1)
=â«logsin2xdx - â«log2dx ...(3)
In the first integral put 2x=t so dx=dt/2
when x=0,t=0 when x=Ï/2 ,t=Ï
So, the first integral can be solved as below:
Now, solve this integral and get the value of the integral I' from the equation (3). Thereafter, equate the value of I' in (A) to get the value of the required integral I.