I used substituition as required to solve this: Let $t = \frac{1}{2}\cos^2\theta$ where $0 \leq \theta \leq \frac\pi2$.
So I want to solve $$\int \dfrac{\sqrt{t}}{1-2t}dt$$ By substituition, we have this form $$\dfrac{1}{\sqrt{2}}\int\csc\theta - \sin\theta ~d\theta$$
The end result is $$\frac{1}{\sqrt{2}}\left[-\ln\left|\csc\theta+\cot\theta\right| + \cos\theta + C\right]$$
Now I have some trouble subbing in $\theta = \cos^{-1}\sqrt{2t}$ back to the equation.
On
$$\int {\sqrt{t}\over 1-2t}dt\tag1$$
$t={1\over 2}\cos^2 \theta\tag a$
$\sqrt{t}={\sqrt{2}\over 2}\cos \theta \tag b$
$1-2t=1-\cos^2 \theta=\sin^2 \theta \tag c$
${dt\over d\theta}=-\sin\theta \cos\theta\tag d$
$$-{\sqrt{2}\over 2}\int {\cos^2 \theta\over \sin \theta }d\theta \tag2$$
Standard integral formula sheet:
$$\int {\cos^2 a\theta\over \sin a\theta}d \theta={\cos a\theta\over a}+{1\over a}\ln \tan{a\theta \over 2}\tag3$$
setting $a=1$
$$-{\sqrt{2}\over 2}\int {\cos^2 \theta\over \sin \theta }d\theta=\cos \theta+\ln\tan {\theta \over 2} \tag4$$
$\tan {\theta \over 2}={1-\cos \theta \over \sin \theta}={\sin \theta\over 1+\cos\theta }\tag5$
$$\int {\sqrt{t}\over 1-2t}dt=-\sqrt{t}+{\sqrt{2}\over 2}\ln\left( \sqrt{1-2t}\over {1-\sqrt{2t}}\right)+c\tag6$$
Hint:
Let $\sqrt t=v\implies t=v^2,dt=2v\ dv$
$$\int\dfrac{\sqrt t\ dt}{1-2t}=\int\dfrac{2v^2\ dv}{1-2v^2}$$
$$=\int\dfrac{(2v^2-1+1)\ dv}{1-2v^2}=\dfrac12\int\dfrac{dv}{(1/\sqrt2)^2-v^2}-\int dv$$