Use the substitution \(u = 1 + \sin ^ { 2 } x\) to show that
$$\int _ { 0 } ^ { \frac { \pi } { 6 } } \frac { 8 \tan x } { 1 + \sin ^ { 2 } x } \mathrm {~d} x = \int _ { p } ^ { q } \frac { 4 } { u ( 2 - u ) } \mathrm { d } u$$
where \(p\) and \(q\) are constants to be found.
Hence, using algebraic integration, show that
$$\int _ { 0 } ^ { \frac { \pi } { 6 } } \frac { 8 \tan x } { 1 + \sin ^ { 2 } x } \mathrm {~d} x = \ln A$$
where \(A\) is a rational number to be found. [0pt]
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