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int(dx)/((1-tanx))=?...

`int(dx)/((1-tanx))=?`

A

`(1)/(2)log|sinx-cosx|+C`

B

`(1)/(2)x+(1)/(2)log|sinx-cosx|+C`

C

`(1)/(2)x-(1)/(2)log|sinx-cosx|+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`I=int(cosx)/((cosx-sinx))dx=int(-cosx)/((sinx-cosx))dx`.
`=((sinx-cosx)-(sinx+cosx))/(2(sinx-cosx))dx=int{(1)/(2)-(1)/(2)*((sinx+cosx))/((sinx-cosx))}dx`
`=(1)/(2)x-(1)/(2)log|sinx-cosx|+C`.
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  • int((1-tanx)/(1+tanx))^(2)dx=

    A
    `-(tan((pi)/(4)-x)+x)+c`
    B
    `-(tan((pi)/(4)+x)+x)+c`
    C
    `tan((pi)/(4)-x)+x+c`
    D
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    A
    `(1)/(3)log[(cosx-sinx)]^(3)+c`
    B
    `tan(x-(pi)/(4))-x+c`
    C
    `tan((pi)/(4)+x)-x+c`
    D
    `sec^(2)((pi)/(4)+x)+c`

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