Integrating $\sec^2(2x)$: A Puzzler

[karush]

\begin{align*}\displaystyle
I_{7}&=\int_{0}^{\pi/8}\frac{\sec^2(2x)}{2+\tan\left({2x}\right)} \\
&=
\end{align*}
not sure of the u substitution here... if $u=tan(2x)$ then $du=2sec^2(2x)$ but stuck with 2 in the denominator after subst

 

[Greg]

What's the derivative of 2?

 

[Prove It]

So maybe the substitution $\displaystyle \begin{align*} u = 2 + \tan{(2\,x)} \end{align*}$ might be more appropriate...

 

[karush]

$u=2+\tan(2x) \therefore du = 2 \sec^2(2x) \, dx$
so then
$\displaystyle I= 2\int_{2}^{3} \frac{1}{u}\,du=
\frac{1}{2}\left[ln(u)\right]_2^3$
so then
$I=\frac{1}{2}\ln\left({\frac{3}{2}}\right)$
hopefully

 

[Greg]

$$I=\color{red}\frac12\color{black}\int_2^3\frac1u\,\text{ d}u=\left[\frac12\log(u)\right]_2^3$$

Otherwise ok.