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

Question

`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)In (sinx)dx, I_(2)=int_(-pi//4)^(pi//4)In(sinx+cosx)dx`. Then
A. `I_(1)=2I_(2)`
B. `I_(2)=2I_(1)`
C. `I_(1)=4I_(2)`
D. `I_(2)=4I_(1)`

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 - A
`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)`