`I_1=int_0^(pi/2)ln(sinx)dx ,I_2=int_(-pi/4)^(pi/4)ln(sinx+cosx)dxdot` Then (a)`I_1=2I_2` (b) `I_2=2I_1` (c)`I_1=4I_2` (d) `I_2=4I_1`

Options

A. `I_(1)=2I_(2)`
B. `I_(2)=2I_(1)`
C. `I_(1)=4I_(2)`
D. `I_(2)=4I_(1)`

Answer

A

Explanation

`I_(2)=int_(-pi//4)^(pi//4) In(sinx+cosx)dx`
`=int_(0)^(pi//4) (In(sinx+cosx)+In(sin(-x)+cos(-x)))dx`
`=int_(0)^(pi//4)(In(sinx+cosx)+In(cosx-sinx))dx`
`=int_(0)^(pi//4) In(cos^(2)x-sin^(2)x)dx`
`=int_(0)^(pi//4) In(cos2x)dx`
PUtting `2x=t,` and `(dt)/2=dx`, we get
`I_(2)=1/2 int_(0)^(pi//2) In(cost)dt`
`=1/2 int_(0)^(pi//2) In("cos"((pi)/2-t))dt`
`=1/2int_(0)^(pi//2) In(sint)dt=1/2 I_(1)`
or `I_(1)=2I_(2)`