# Simple integration by parts problem?

An integration reduction formula for $\sec^n x$ is

$$\int\sec^nx\;dx=\frac{1}{n-1}\sec^{n-2}x\tan x+\frac{n-2}{n-1}\int\sec^{n-2}x\;dx, n≠1$$

Using this formula (which I am sure is correct) gives the integral of $16\sec^3x$ to be

$$16\int \sec^3 x\;dx=16\left(\frac{1}{2}\sec x\tan x-\frac{1}{2}\int\sec x\;dx\right)=8\sec x\tan x-8\ln|\tan x+\sec x|+C$$

Now, if I don't use this reduction formula, and instead opt to do it using the same method used to generalize the formula in the first place (integration by parts) I should get the same answer. But I don't! Aha! There lies my dilemma. Watch:

Letting $u=\sec x$ and $dv=\sec^2x$, then $du=\tan x\sec x$ and $v=\tan x$ and we get

$$\int{\sec^3xdx}=\sec x\tan x-\int\sec x\tan^2x\;dx$$

Using the Pythagorean trigonometric identity $\tan^2x=\sec^2x-1$

$$\int\sec^3x\;dx =\sec x\tan x-\int \sec x(\sec^2x-1)\;dx =\sec x\tan x-\int\sec^3x-\sec x\;dx$$

And so, using the integral used earlier, we get

$$16\int\sec^3x\;dx=16(\sec x\tan x-\int \sec^3x\;dx +\ln|\tan x+\sec x|)$$

However, if I was to multiply the $16$ on the RHS "in" and then absorb the integral on the LHS and divide, I would end up with

$$\int\sec^3x \;dx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\tan x+\sec x|+C$$

which isn't the same answer I arrived at earlier. What am I doing wrong? I'm sure that it's probably something very simple, but I cannot figure it out. I apologize in advance if maybe this question is too simple to be asked here. Thanks in advance.

The reduction formula (which you were sure about :)) has a typo (edit: now corrected). The correct formula is: $$\int{\sec^nx}dx=\frac{1}{n-1}\sec^{n-2}x\tan x+\frac{n-2}{n-1}\int{\sec^{n-2}xdx},$$ for $n \geq 2$. (Note that the second term should carry a "$+$" sign and the exponent of $\sec x$ is $n-2$.) You can confirm the formula and find its derivation here.

Also, as Brian points out, if you want $16 \int \sec^3 x dx$ rather than $\int \sec^3 x dx$ (which is what both the methods are really calculating), you should multiply the answer by $16$. I will record the correct answer here for easy reference: $$\int 16\ \sec^3 x dx = 8 \sec x \tan x + 8 \ln |\sec x + \tan x| + C.$$

• Yes, you are correct about the formula being wrong, I have now edited my post to reflect this. However, I still don't see what I did wrong. Could you explain what you mean by having to multiply by 16? Sep 21, 2011 at 5:59
• Ok. See, you did nothing wrong in the second method; in particular, the correct answer is in the very last equation. Now, your very first equation which is supposed to be a general formula to integrate $\sec^n x$, has a typo. Moreover, you multiplied that formula by 16 for some reason that I don't understand. Can you clarify what the exact integrand is: $\sec^3 x$ or $16\sec^3 x$? @Haut Sep 21, 2011 at 6:03
• I got it now. Thank you for your help. Sep 21, 2011 at 6:10
• Ok, the correct answer is: $$16 \int \sec^3 x = 8 \sec x \tan x + 8 \ln |\tan x + \sec x| + C.$$ If you correct the original formula (change the - sign to + and also change the exponent), then you will get the exact same answer. @Haut Sep 21, 2011 at 6:11

It's not the same answer because it’s not the same question: your first result was for $16\int \sec^3 x dx$, your second for $\int \sec^3 x dx$. Multiply your second result by $16$, and you’ll be fine.

Added: Well, you would have been fine if you’d got the sign right in the reduction formula, which I carelessly failed to check. See Srivatsan’s answer for details. (I’m leaving mine up on account of the note below.)

Technical note: In your integration by parts, $dv$ is not $\sec^2 x$: it’s $\sec^2 x dx$. Similarly, $du$ is not $\tan x \sec x$ but rather $\tan x \sec x dx$.

• Actually, the answers are not the same even after multiplying by 16. The second term in the very first formula is negative, but it seems it should be positive. Sep 21, 2011 at 5:44
• @Srivatsan: That’s what I get for not double-checking the part of which he was certain. (The $n-1$ exponent ought to have been a dead giveaway that it needed a second look.) sigh Sep 21, 2011 at 6:00
• I integrated $\int{sec^3x}dx$ by parts, and then I multiplied all of that by 16. I'm sure the typos had nothing to do with the answers not being the same since I did this on paper with the proper formula, I just mistyped it. Could you help me understand what you mean? Sep 21, 2011 at 6:05
• @Brian Actually even I took that on faith. Only after checking that the signs on the right hand side did not match (and noting that the integration by parts approach looks ok), did I suspect that something is off with the formula. Sep 21, 2011 at 6:06
• @Hautdesert: You multiplied the result of integrating $\int\sec^3 x dx$ by parts by $16$, but then you solved for $\int\sec^3 x dx$ instead of for $16\int\sec^3 x dx$. Your original result using the reduction formula was for $16\int\sec^3 x dx$. Sep 21, 2011 at 6:22