Question: My attempt:

$$\eqalign{
 & \int\_0^{\pi \over 3} \sec x\tan x\sqrt {\sec x + 2} \, dx \cr 
 & u = \sec x \cr 
 & {du \over dx} = \sec x\tan x \cr 
 & {dx \over du} = {1 \over \sec x\tan x} \cr 
 & \int\_0^{\pi \over 3} \sec x\tan x\sqrt {\sec x + 2} \, dx = \int\_1^2 {\sec x\tan x\sqrt {u + 2} \over \sec x\tan x} \, du \cr 
 & = \int\_1^2 \sqrt {u + 2} \, du \cr 
 & = \int\_1^2 (u + 2)^{1 \over 2} \, du \cr 
 & = \int\_1^2 u^{1 \over 2} + \sqrt 2 \, du \cr 
 & = \left[ {{2 \over 3}{u^{{3 \over 2}}} + u\sqrt 2 } \right]\_1^2 \cr 
 & = \left[ {2 \over 3}(2)^{3 \over 2} + 2\sqrt 2 \right] - \left[ {2 \over 3}(1)^{3 \over 2} + \sqrt 2 \right] \cr 
 & = 2.633\ldots \cr} $$

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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!

Answer: 
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 $\sqrt{u+2}\ne \sqrt{u}+\sqrt{2}$.
$ \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 $v=u+2$…