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.

  • 1
    $\begingroup$ 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

If $n$ is a whole number, we can use integration by parts to find the general integral $$I=\int \sec^nx\ dx$$ $$I=\int\sec^{n-2}x\sec^2x\ dx$$ $dv=\sec^2x\ dx$

$v=\tan x$


$du=(n-2)\sec^{n-2}x\tan x\ dx$ $$I=uv-\int vdu=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x\tan^2x\ dx$$ $$I=\sec^{n-2}x\tan x-(n-2)\int\sec^{n-2}x(\sec^2x-1)dx$$ $$I=\sec^{n-2}x\tan x-(n-2)\int\sec^nx\ dx+(n-2)\int\sec^{n-2}x\ dx$$ $$I=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}x\ dx-(n-2)I$$ $$I+(n-2)I=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}x\ dx$$ $$(n-1)I=\sec^{n-2}x\tan x+(n-2)\int\sec^{n-2}x\ dx$$ $$I=\frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}x\ dx$$ 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 $\sec x$.


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

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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.