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int(0)^(pi//2)(tanx)/(1+m^(2)tan^(2)x)\ ...

`int_(0)^(pi//2)(tanx)/(1+m^(2)tan^(2)x)\ dx`

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The correct Answer is:
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Let `I = int_(0)^(pi//2) (tanxdx)/(1+m^(2)tan^(2)x)dx`
`= int_(0)^(pi//2) ((sinx)/(cosx))/(1+m^(2).(sin^(2)x)/(cos^(2)x))dx`
`= int_(0)^(pi//2)((sinx)/(cosx))/((cos^(2)x+m^(2)sin^(2)x)/(cos^(2)x))dx`
`= int_(0)^(pi//2)(sinxcosxdx)/(1-sin^(2)x+m^(2)sin^(2)x)dx`
` = int_(0)^(pi//2)(sinxcosx)/(1-sin^(2)x(1-m^(2)))dx`
Put `sin^(2)x=t`
`rArr 2sinx cosx dx = dt`
`:. I = 1/2int_(0)^(1)(dt)/(1-t(1-m^(2)))`
`= 1/2[-log|1-t(1-m^(2))|.(1)/(1-m^(2))]_(0)^(1)`
`= 1/2[-log|1-1+m^(2)|.(1)/(1+m^(2))+log|1|.(1)/(1+m^(2))]`
`=1/2[-log|m^(2)|.(1)/(1-m^(2))]=2/(2).(logm)/((m^(2)-1))`
`log'(m)/(m^(2)-1)`
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