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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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    Ya youre right :)2010-12-05

3 Answers 3

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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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    OK steps posted2010-12-05
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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.