# Trigonometric integral question

The problem was to find $\int \tan^5(x)\, \sec^7(x)\, dx$

The solution the book got was different from mine, but I'm sure I did the right steps. Since the solutions are completely different, I have a feeling they might be the same thing, but I'm not sure how I would know this. Here's what the book got:

$\sec^{11}(x)/11-2\sec^9(x)/9+\sec^7(x)/7+C$

Here's what I got: $\tan^6(x)/6 + \tan^{11}(x)/11 + C$

EDIT: So my answer is wrong then, so let me list my steps see if you can catch my mistake

$\int \tan^5(x)\sec^7(x)\, dx$

$\int \tan^4x \tan x \sec^5x \sec^2x\, dx$

$\int \tan^4 x\tan x \ (1+\tan^5x)\,\sec^2x\,dx$

$u=\tan x \ du=\sec^2x$

$\int u^4(u)(1+u^5)\,du$

$\int u^5(1+u^5)\,du$

$\int u^5+u^{10}\,du$

$=\tan^6(x)/6 + \tan^{11}(x)/11 + C$

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$tan^5(x) + 1 \neq sec^5(x)$. –  Zarrax Dec 5 '10 at 22:43
Ya youre right :) –  maq Dec 5 '10 at 23:14

If they differ by a constant then their difference should be constant. You can check that this is not so by setting $x=0$ and $x=\pi.$

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The difference of what? Both equations? I'm not understanding what you mean –  maq Dec 5 '10 at 22:03
@f-prime The first answer minus the second. Note that the first answer switches sign when we put in $x=\pi$ instead of $x=0.$ –  Derek Jennings Dec 5 '10 at 22:07
The difference of the two solutions. At zero, your solution is C, while the book's solution is 1/11-2/9+1/7+C'. At pi, your solution is again C, while the book's is -(1/11-2/9+1/7)+C'. C'-C is not consistent between these, so they differ. –  Ross Millikan Dec 5 '10 at 22:07
But if I plug both of these in Wolframalpha for say x=5, the book solution comes out to .47 and my solution comes out to 7.47, so they differ by a constant of 7, no? Or am I misunderstanding the concept of differing by a constant? –  maq Dec 5 '10 at 22:11
Ignoring $C$ the second is $0$ for both $x=0$ and $x=\pi$ but the first changes from $8/693$ to $-8/693$ so their difference cannot be constant. –  Derek Jennings Dec 5 '10 at 22:11

Instead of working with $\tan$ ,$\sec$ , use their $\sin \text {and} \cos$ definitions in their place, the problem will be reduced to reverse application of product rule. way easier than to play around with than $\tan$ and $\sec$ .

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Here's a hint: calculate the derivative of $\sec^n(x)$ for integer n > 0, then think how you can use this , and the fact that $\tan^2(x) = 1 + \sec^2(x)$, to calculate the integral.

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