How do I solve for k in an equation with integrals?

Question

$\int_0^k \frac{sec^2(x)}{1 + tan(x)} = ln(2)$

I need to solve the above equation for k. This is what I did:

$[ln|1+tan(x)|]_0^k = ln(2)$
$ln|1 + tan(k)| - ln|1 + tan(0)| = ln(2)$
$ln|1 + tan(k)| = ln(2)$
$|1 + tan(k)| = 2$
$1 + tan(k) =+-2$
$tan(k) = -1+-2$

This gives us:

$k = pi/4 + npi$ where n is any integer

and

$k = tan^-1(-3) + npi$ where n is any integer

My problem is that the only value of k that actually works in the original equation is $k = pi/4$ .

Why do the other ones not work (for example $k = (5pi)/4$ )?

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Answer 1 · 2017-08-15T21:54:21

If you try to integrate past an asymptote, the integral does not converge.

The integrand is not defined where $tanx = -1$ . Nor at $x = pi/2+pik$ for integer $k$ .

And the improper integral past such a point does not converge.