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

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?

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

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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.

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  • $\begingroup$ Wow, for some reason the remaining $sec(x)$ was just not clicking to also sub that for $u$. $\endgroup$ – DJ2 Sep 25 '18 at 2:58

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