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My attempt:

$$\eqalign{ & \int_0^{\pi \over 3} \sec x\tan x\sqrt {\sec x + 2} \, dx \cr & u = \sec x \cr & {du \over dx} = \sec x\tan x \cr & {dx \over du} = {1 \over \sec x\tan x} \cr & \int_0^{\pi \over 3} \sec x\tan x\sqrt {\sec x + 2} \, dx = \int_1^2 {\sec x\tan x\sqrt {u + 2} \over \sec x\tan x} \, du \cr & = \int_1^2 \sqrt {u + 2} \, du \cr & = \int_1^2 (u + 2)^{1 \over 2} \, du \cr & = \int_1^2 u^{1 \over 2} + \sqrt 2 \, du \cr & = \left[ {{2 \over 3}{u^{{3 \over 2}}} + u\sqrt 2 } \right]_1^2 \cr & = \left[ {2 \over 3}(2)^{3 \over 2} + 2\sqrt 2 \right] - \left[ {2 \over 3}(1)^{3 \over 2} + \sqrt 2 \right] \cr & = 2.633\ldots \cr} $$


This is the wrong answer, the right answer is ${{16} \over 3} - 2\sqrt 3 $ (1.87 in decimal form).

Could someone explain where I went wrong please?

Thank you!

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    $\begingroup$ $(u+2)^{\frac{1}{2}} \ne u^{\frac{1}{2}} + \sqrt{2} $ $\endgroup$ – Random Variable May 10 '13 at 0:33
  • $\begingroup$ Please don't use \limits on the title of a question: it doesn't render prettily on the main page. Thanks. $\endgroup$ – Bruno Stonek May 10 '13 at 0:38
  • $\begingroup$ @RandomVariable I see... is it possible to expand this out at all? Furthermore, if I were to integrate this I'd have to do it as a function right? $\endgroup$ – seeker May 10 '13 at 0:38
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    $\begingroup$ Make another substitution $v=u+2$. Or make the initial substitution that André Nicolas suggests below. $\endgroup$ – Random Variable May 10 '13 at 0:40
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Let $u=\sec x+2$. Then $du=\sec x\tan x\,dx$, so we are finding $\displaystyle\int_{u=3}^4 u^{1/2}\,du$.

Remark: Substitution is a much simpler technique than you make it to be. The substitution $u=\sec x$ also works nicely, in pretty much the same way, except that we end up integrating $\sqrt{u+2}\,du$. You got to that, after more time than necessary, and then made an algebra slip. Of course you know that $\sqrt{u+2}\ne \sqrt{u}+\sqrt{2}$.

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  • $\begingroup$ Could you explain how I spent more time on it than necessary? I'd love to know how I could have done this faster, I really am prone to a lot of errors on other similar questions when I use substitution so your insight would be appreciated. Thanks! $\endgroup$ – seeker May 10 '13 at 0:42
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    $\begingroup$ @Assad: It is contained in my answer. We go directly to $du=\sec x\tan x\,dx$, and see that that is part of the expression we want to integrate (which is what motivated the substitution). So we replace $\sec x\tan x\,dx$ by $du$, replace $\sec x+2$ by $u$, and we are done. I was also recalling an earlier question I am pretty sure was from you (same style) in which substitution took a long time. Don't bother with this $\frac{dx}{du}$ stuff. $\endgroup$ – André Nicolas May 10 '13 at 0:47
  • $\begingroup$ I see! I understand now, thanks! $\endgroup$ – seeker May 10 '13 at 0:53
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$ \int_1^2 {{{(u + 2)}^{{1 \over 2}}}du} \neq \int_1^2 {{u^{{1 \over 2}}} + \sqrt 2 du} $

If this is hard to see let $v=u+2$…

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  • $\begingroup$ I see the mistake, thank you! $\endgroup$ – seeker May 10 '13 at 0:40

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