Prove that `int_(-pi//4)^(pi//4) log(sinx + cosx)\ dx = -pi/4\ log2.`

Answer

N/a

Explanation

Let `I = int_(-pi//4)^(pi//4)log(sinx+cos)dx"....."(i)`
`I = int_(-pi//4)^(pi//4)log{sin(pi/4-(pi)/(4)-x)+cos((pi)/(4)-(pi)/(74)-x)}dx`
`= int_(-pi//4)^(pi//3)log{sin (-x)+ cos(-x)}+dx`
`= int_(-pi//4)^(pi//4)log(cosx-sinx)dx"....."(ii)`
From Eqs. (i) and (ii),
` 2I = int_(-pi//4)^(pi4)logcos2xdx`
`2I = int_(0)^(pi//4)logcos2x dx"....."(iii)`
`[:' int_(-a)^(a)f(x)= 2 int_(0)^(a) f(x),"if" f(-x) = f(x)]`
Put `2x = t rArr dx = (dt)/(2)`
As `x rarr 0`, then `t rarr 0`
and `x rarr pi/4`, then `t rarr pi/2`
`2I = 1/2 int_(0)^(pi//2)logcost dt "...."(iv)`
`2I = 1/2 int_(0)^(pi//2)log cos(pi/2 - t) dt , [:' int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`rArr 2I = 1/2 int_(0)^(pi//2)logsintdx"........."(v)`
On adding Eqs. (iv) and (v), we get
`4I = 1/2 int_(0)^(pi//2) log sint cost dt`
`rArr 4I = 1/2 int_(0)^(pi//2)log'(sin2t)/(2)dt`
`4I = 1/2 sin_(0)^(pi//2)logsin2xdx - 1/2 int_(0)^(pi//2) log 2dx`
`rArr 4I = 1/2 int_(0)^(pi//2) logsin(pi/2 - 2x) dx - log2.(pi)/(4)`
`rArr 4I = 1/2 int_(0)^(pi//2)log cos2x dx - (pi)/(4) log2`
`rArr 4 I = int_(0)^(pi//4) logcos2xdx - (pi)/(4) log2 [:' int_(0)^(2a)f(x)dx=2int_(0)^(a) f(x)dx]`
`rArr 4I = 2I - (pi)/(4) log2` [from Eq. (iii)]
`:. I = - (pi)/(8) log2 = (pi)/(8) log(1/2)`