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?

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 $$\int\tan x\cdot2\sqrt{\sec x}\ dx=4\sqrt{\sec x}$$ $$=\ln(\sec x)\cdot2\sqrt{\sec x}-4\sqrt{\sec x}+C$$
Yes! The $$\tan(x)$$ cancels.
I'm sure you know what to do with the remaining $$\sec(x)$$, given that $$u=\sec(x)$$.
• Wow, for some reason the remaining $sec(x)$ was just not clicking to also sub that for $u$. – DJ2 Sep 25 '18 at 2:58