Using the substitution u = tan x, find the exact value of ∫_(π/6)^(π/3) 1/(1 - sin² x) dx.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. M1: for stating du = sec² x dx or du/dx = sec² x.
2. B1: for recognizing that 1 - sin² x = cos² x.
3. The integral becomes ∫ 1/(cos² x) dx = ∫ sec² x dx.
4. M1: for correctly changing the limits.
5. When x = π/6, u = tan(π/6) = 1/√3.
6. When x = π/3, u = tan(π/3) = √3.
7. The integral becomes ∫_(1/√3)^(√3) sec² x dx. (This step implicitly uses dx = du/(sec² x), so ∫ sec² x du/(sec² x) = ∫ du)
8. A1: for correctly transforming the integral into ∫_(1/√3)^(√3) 1 du (or ∫ du).
9. M1: for integrating 1 with respect to u to get [u].
10. A1: for evaluating the definite integral correctly: [u]_(1/√3)^(√3) = √3 - 1/√3 = (3-1)/√3 = 2/√3 = 2√3/3.