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I can't find a way to evaluate this: $$\int x\sec(x)\tan(x)\:\mathrm{d}x$$ I'm not sure how i have to proceed. I've tried the substitution $u=\tan(x)$, but it got me nowhere.

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  • $\begingroup$ what's tg(x)? Can you give the definition? $\endgroup$ – voldemort Apr 6 '14 at 14:34
  • $\begingroup$ it's tan, sin/cos $\endgroup$ – Wollacy Silva Apr 6 '14 at 14:35
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Use integration by parts:

$$\int u\;dv = uv - \int v\;du$$

Let $u = x\implies du = dx$.

And let $dv = \sec x \tan x \implies v = \sec x$.

$$\int x\,\sec x \tan x \;dx = x \,\sec x - \int \sec x\,dx$$

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Notice:

$$ \int x \sec x \tan x dx = \int x d( \sec x ) = x \sec x - \int \sec x dx $$

Using integration by parts. Now to integrate $\int \sec x $, notice

$$ \int \sec x = \int \sec x \frac{ ( \tan x - \sec x ) }{(\tan x - \sec x)} dx = \int \frac{\sec x \tan x - \sec^2 x}{\tan x - \sec x } dx = -\int \frac{d(\sec x - \tan x)}{\sec x - \tan x} = \; \; \; \; -\ln ( \sec x - \tan x ) + C$$

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  • $\begingroup$ How that: ∫(sec^2(x)−sec(x)tan(x)dx)/sec(x)-tan(x) turns into: ∫(d(secx−tanx))/secx−tanx ? $\endgroup$ – Wollacy Silva Apr 6 '14 at 14:47
  • $\begingroup$ editted the answer $\endgroup$ – user139708 Apr 6 '14 at 14:51

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