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.
 A: 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: 
$$
\int 16\ \sec^3 x dx = 8 \sec x \tan x + 8 \ln |\sec x + \tan x| + C.
$$
A: 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 $\tan x \sec x dx$.
