Easiest way to evaluate $\int_0^{\pi/6}\sec x\,dx$? Per the question title, what's the easiest way to evaluate the following?
$$\int_0^{\pi/6}\sec x\,dx$$
You can do something like computing the derivatives of $\sec x$ and $\tan x$, adding them up, computing the derivative of the logarithm of the absolute value of the sum of $\sec x$ and $\tan x$, and then completing the integration-by-parts, getting the final answer of $\ln(\sqrt{3})$.
But that feels like pulling something out of thin air.
I'm wondering if there's an easier way to compute the integral.
 A: Let's make the famous $z=\tan(\frac{x}{2})$ substitution. Then $\sec x=\frac{1+z^2}{1-z^2}$ and $dx=\frac{2dz}{1+z^2}$. Knowing that $\tan(\frac{\pi}{12})=2-\sqrt{3}$, we compute $$\int_0^{\frac{\pi}{6}}\sec{x}dx=\int_0^{2-\sqrt{3}}\frac{-2dz}{z^2-1}=\ln\left (\left|\frac{z+1}{z-1}\right|\right )|_0^{2-\sqrt{3}}=\ln\left(\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\ln(\sqrt{3}).$$
A: It's not difficult (and a standard exercise) to compute
$$
\int\frac{2}{\sin2t}\,dt=\int\frac{\cos^2t+\sin^2t}{\sin t\cos t}\,dt=
\int\Bigl(\frac{\cos t}{\sin t}+\frac{\sin t}{\cos t}\Bigr)\,dt=\log\lvert\tan t\rvert+c
$$
How do you transform a cosine into sine? Easy, with the complementary angle. So perform $x=\pi/2-2t$ and your integral becomes
$$
\int_{\pi/4}^{\pi/6} -\frac{2}{\sin2t}\,dt=\Bigl[\log\tan t\Bigr]_{\pi/6}^{\pi/4}=-\log(1/\sqrt{3})=\log\sqrt{3}
$$
A: $$\int \sec x\,dx= \int \frac{\sec^2x}{\sqrt{1+\tan^2x}}dx=\sinh^{-1}(\tan x)+C
$$
A: $$\int_0\frac{1}{1-u^2}du=\text{artanh}(u)$$
$$\int_0\frac{1}{\cos x}dx=\int_0\frac{1}{1-\sin^2 x}\cos x\,dx=\text{artanh}(\sin x)$$
$$\int_0^{\pi/6}\frac{1}{\cos x}dx=\text{artanh}(\sin(\pi/6))=\text{artanh}(1/2)$$
