Finding $\int\sqrt{\sec(x)}\ln(\sec(x))\tan(x)\,\mathrm dx$ by substitution $$\int\sqrt{\smash[b]{\sec(x)}}\ln(\sec(x))\tan(x)\,\mathrm{d}x$$


I started by making u-substitution $u = \sec(x)$:

$$\int\sqrt u\ln(u)\tan(x) \left(\frac{\mathrm{d}u}{\sec(x)\tan(x)}\right)$$

Now, does the $\tan(x)$ cancel? Then is integration by parts the appropriate method to use?
Yes! The $\tan(x)$ cancels.

I'm sure you know what to do with the remaining $\sec(x)$, given that $u=\sec(x)$.

Integration by parts is next.
Given $$\int\sqrt{\sec x}\ln(\sec x)\tan x\ dx$$
Apply integration by parts $u=\ln(\sec x),v^{\prime}=\sqrt{\sec x}\tan x$

$$=\ln(\sec x)\cdot2\sqrt{\sec x}-\int\tan x\cdot2\sqrt{\sec x} dx$$
Since $\int\tan x\cdot2\sqrt{\sec x}\ dx=4\sqrt{\sec x}$
$$=\ln(\sec x)\cdot2\sqrt{\sec x}-4\sqrt{\sec x}+C$$