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I1=int0^(pi/2)ln(sinx)dx ,I2=int(-pi/4)^...

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

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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 I_1=4I_2 (d) I_2=4I_1

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 I_1=4I_2 (d) I_2=4I_1

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

I_(1)=int_(0)^((pi)/2)Ln (sinx)dx, I_(2)=int_(-pi//4)^(pi//4)Ln(sinx+cosx)dx . Then

I_(1)=int_(0)^((pi)/2)In (sinx)dx, I_(2)=int_(-pi//4)^(pi//4)In(sinx+cosx)dx . Then

I_(1),=int_(0)^((pi)/(2))1n(sin x)dx,I_(2)=int_(-(pi)/(4))^((pi)/(4))1n(sin x+cos x)dx. Then I_(1)=2I_(2)(b)I_(2)=2I_(1)I_(1)=4I_(2)(d)I_(2),=4I_(1)

If I_(1) = int_(0)^(pi//2)ln (sin x)dx , I_(2)=int_(-pi//4)^(pi//4)ln (sin x + cos x)dx , then :

If I_1=int_0^(pi/2)f(sinx)sinxdx and I_2=int_0^(pi/2)f(cosx)cosxdx then I_1/I_2

If I_1=int_0^(pi/2)f(sinx)sinxdx and I_2=int_0^(pi/2)f(cosx)cosxdx then I_1/I_2

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