`int_(pi//6)^(pi//3)(2\ dx)/(1+sin\ x)`

We need to re-express this and we make use of a fairly common technique. We multiply the numerator (top) and denominator (bottom) of the fraction by the conjugate of the denominator. (The conjugate has the opposite sign in the middle. In this case, it's a minus sign.)

We will make use of this important result:

`sin^2 x+ cos^2 x = 1`

`1/(1+sin\ x) =1/(1+sin\ x)xx(1-sin\ x)/(1-sin\ x)`

`=(1-sin\ x)/(1-sin^2x)`

`=(1-sin\ x)/(cos^2 x)`

`=sec^2x-(sin\ x)/(cos^2 x)`

Now we can integrate.

Putting `u = cos\ x` in the right hand part, we have:

`du = -sin\ x\ dx`

So

`int_(pi//6)^(pi//3)(2\ dx)/(1+sin\ x)`

`=2int_(pi//6)^(pi//3)(sec^2x-(sin\ x)/(cos^2x))dx`

`=2[tan\ x-1/(cos\ x)]_(pi//6)^(pi//3)`

`=2[(1.73215-2)-` `{:(0.57735-1.15470)]`

`=0.6190`

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