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Evaluate intpi/6pi/3(dx)/(1+sqrt(tanx...

Evaluate `intpi/6pi/3(dx)/(1+sqrt(tanx))""`

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Denoted by `I=int_((pi)/(6))^((PI)/(3))(dx)/(1+sqrt(tanx))=int_((pi)/(6))^((pi)/(3))(sqrt(cosx))/(sqrt(cosx)+sqrt(sinx))dx`……..`(i)`
Now `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`,
Then `l=int_((pi)/(6))^((pi)/(3))(sqrt(cos((pi)/(3)+(pi)/(6)-x)))/(sqrt(cos((pi)/(3)+(pi)/(6)-x)+sqrt(sin((pi)/(3)+(pi)/(6)-x))))dx`
`=int_((pi)/(6))^((pi)/(3))(sqrt(cos((pi)/(2)-x)))/(sqrt(cos((pi)/(2)-x)+sqrt(sin((pi)/(2)-x))))dx`
i.e., `l=int_((pi)/(6))^((pi)/(3))(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx`......`(ii)`
Adding `(i)` and `(ii)`, we get
`2l=int_((pi)/(6))^((pi)/(3))(sqrt(sinx)+sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`
`=int_((pi)/(6))^((pi)/(3))dx=(pi)/(3)-(pi)/(6)=(pi)/(6)`
`:.l=(pi)/(12)`
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