3
I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{2}\left [\underbrace{\int\frac{\sqrt{1+x}}{x}}_{J} d x+\und...
0
Multiplying both numerator and denominator of the integrand by $\sec^2 x$ yields
\begin{aligned}
& \int \frac{d x}{1-\sin x \cos x} \\
=& \int \frac{\sec ^{2} x}{\sec ^{2} x-\tan x} d x \\
=& \int \frac{d(\tan x)}{\tan ^{2} x-\tan x+1} \\
=& 2 \int \frac{d(2 \tan x-1)}{(2 \tan x-1)^{2}+(\sqrt{3})...
