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The integral of tan(x) dx = ln | sec(x) | + c
to solve... tan(x) = sin(x)/cos(x)
the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx
= the integral of (1/u -du)
= -ln | u | + c
= -ln | cos(x) | + c
= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c
= ln | sec(x) | + c
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