= Solution
Let $h=\sqrt{1-m^2}$ be the complementary modulus of the <complete elliptic integral of the first kind>. With $\varphi=\pi/2-\theta$, write the integrand as $(\sin^2\varphi+h^2\cos^2\varphi)^{-1/2}$. Its <logarithmic endpoint asymptotic of the complete elliptic integral> comes from $\varphi=O(h)$; an expansion at fixed $\varphi$ alone misses the constant accompanying the logarithm.
Choose an intermediate cutoff $h\ll d\ll1$. In the endpoint region, the integral is
$$
\int_0^d\frac{d\varphi}{\sqrt{\varphi^2+h^2}}=\operatorname{arsinh}(d/h)=\log\frac{2d}{h}+o(1).
$$
Outside this region, the leading integral is
$$
\int_d^{\pi/2}\frac{d\varphi}{\sin\varphi}=-\log\tan(d/2)=\log\frac2d+o(1).
$$
The arbitrary cutoff cancels in this <matched asymptotic expansion>, leaving
$$
\boxed{K(m)=\log\frac4{\sqrt{1-m^2}}+o(1)=-\frac12\log(1-m)+\log(2\sqrt2)+o(1).}
$$
These are the divergent logarithm and the finite constant, the requested first two terms. If terms are instead grouped by powers of the complementary modulus, the next refinement is $K(m)=\log(4/h)+(h^2/4)[\log(4/h)-1]+O(h^4\log(1/h))$; its coefficients agree with https://dlmf.nist.gov/19.12.E1[the DLMF expansion].
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