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If I=int(sqrt(cotx)-sqrt(tanx))dx, then ...

If `I=int(sqrt(cotx)-sqrt(tanx))dx,` then `I` equals

A

`sqrt(2)log(sqrt(tanx)-sqrt(cotx))+C`

B

`sqrt(2)log|sinx+cosx+sqrt(sin2x)|+C`

C

`sqrt(2)log|sinx-cosx+sqrt(2)sinx cosx|+C`

D

`sqrt(2)log|sin(x+pi//4)+sqrt(2)sinx cosx|+C`

Text Solution

Verified by Experts

The correct Answer is:
B
`I=int(cosx-sinx)/(sqrt(cosx sinx))dx`
Put `sinx+cosx=t, " so that " 2sinx cosx=t^(2)-1`
` :. I=sqrt(2)int(dt)/(sqrt(t^(2)-1))=sqrt(2)log|t+sqrt(t^(2)-1)|+C`
`=sqrt(2)log|sinx+cosx+sqrt(sin2x)|+C`
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