`I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx`, `I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1

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`I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx`, `I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1

`I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx`,
`I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1)dx`. Then
A. `I_(2)=I_(3)=I_(4)=0,I_(1)!=0`
B. `I_(1)=I_(2)=I_(3)=0,I_(4)!=0`
C. `I_(1)=I_(3)=I_(4)=0,I_(2)!=0`
D. `I_(1)=I_(2)=I_(3)=0,I_(4)!=0`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
`I_(1)=int_(0)^(pi//2)(sinx_cosx)/(1+sinx cos) dx`
`=int_(0)^(pi//2) ("sin"((pi)/2-x)-"cos"((pi)/2-x))/(1+"sin"((pi)/2-x)"cos"((pi)/2-x))`
`=int_(0)^(pi//2)(cosx-sinx)/(1+sinx cosx)dx=-I_(1)`
or `I_(1)=0`
`I_(3)=0` as `sin^(3)x` is odd
`I_(4)=int_(0)^(1)In((1-x)/x)dx`
`=int_(0)^(1)In((1-(1-x))/(1-x))dx`
`=int_(0)^(1)In x/(1-x)dx=-I_(4)`
or `I_(4)=0`
`I_(2)=int_(0)^(2pi)cos^(6)x dx=2int_(0)^(pi)cos^(6)x dx!=0`