Integrating $\sec^5x$ using integration by parts
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.
$\endgroup$ 32 Answers
$\begingroup$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 $\sec x$.
$\endgroup$ $\begingroup$$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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