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I'd like to know what I did wrong in my solution using this particular $u$ and $v$

$\int\sec^5x dx$

$u=\sec x$

$du=\sec x\tan x$

$v=\int \sec^4x dx = (\tan^3x)/3+\tan x$

After completing the integral, I ended up with $\int\sec^5x dx= \frac{\tan^3x\sec x}4+\frac{3\tan x\sec x}8+\frac{3\ln(\tan x+\sec x)}8+C$

Which is very close to the correct answer but not quite. I can't figure out where I went wrong at all.

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    $\begingroup$ What "correct" answer are you aiming for? Can you differentiate the two answers and compare the resulting derivatives? Also title of q seems mistyped. $\endgroup$ – Oscar Lanzi Nov 9 '18 at 21:46
  • $\begingroup$ @user3506217 I've made some typographical improvements. Please double-check my edit in case I misunderstood any of you formulae. $\endgroup$ – J.G. Nov 9 '18 at 22:22
  • $\begingroup$ It's not easy, if at all possible, to determine where the error occurred without seeing more intermediate steps. $\endgroup$ – Travis Willse Nov 9 '18 at 22:36
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If $n$ is a whole number, we can use integration by parts to find the general integral $$I=\int \sec^nx\ dx$$ $$I=\int\sec^{n-2}x\sec^2x\ dx$$ $dv=\sec^2x\ dx$

$v=\tan x$

$u=\sec^{n-2}x$

$du=(n-2)\sec^{n-2}x\tan x\ dx$ $$I=uv-\int vdu=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x\tan^2x\ dx$$ $$I=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x(\sec^2x-1)dx$$ $$I=\sec^{n-2}x\tan x-(n-2)\int\sec^nx\ dx+(n-2)\int\sec^{n-2}x\ dx$$ $$I=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}x\ dx-(n-2)I$$ $$I+(n-2)I=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}x\ dx$$ $$(n-1)I=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}x\ dx$$ $$I=\frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}x\ dx$$ This is called a reduction formula. Note that it does not work for $n=1$.

Just plug in your $n$. I trust that you can integrate $\sec x$.

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$I = \int \sec^5 x \ dx\\ u = \sec^3 x, dv = \sec^2 x\ dx\\ du = 3\sec^2 x(\sec x\tan x)\ dx, v = \tan x$

$\sec^3x\tan x - 3\int \sec^3 x\tan^2 x \ dx\\ \sec^3x\tan x - 3\int \sec^3 x(sec^2 x - 1)\ dx\\ I = \sec^3x\tan x + 3\int \sec^3 x\ dx - 3I\\ 4I = \sec^3x\tan x + 3\int \sec^3 x\ dx\\ I = \frac 14(\sec^3x\tan x + 3\int \sec^3 x\ dx)$

But now we need to do something very similar to find

$J = \int \sec^3 x\ dx\\ J = \sec x\tan x + \int \sec x\ dx + J\\ J = \frac12(\sec x\tan x + \ln|\sec x + \tan x|)$

$\frac 14\sec^3x\tan x + \frac38 \sec x\tan x + \frac 38 \ln|\sec x + \tan x| + C$

How does this compare to:

$\frac{\tan^3 x\sec x}4+\frac{3\tan x\sec x}8+\frac{3\ln(\tan x+\sec x)}8$

We agree on the last two terms, and $\tan^3 x\sec x = \sec^3 x\tan x + \sec x$

So, we differ on a $\frac {\sec x}{4}$ term

Without seeing your work, I can't tell you where that term might have come from (or gone).

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