Problem source

Show that $\cos 4u=8\cos^{4}u-8\cos^{2}u+1$.
If
\[
I=\int_{-1}^{1}
\frac{1}{\vphantom{{\big(}^2}\; 
\surd(1+x)+\surd(1-x)+2\;
}\;{\rm d}x  ,\]
show, by using the change of variable $x=\cos t$, that
\[
I= \int_0^\pi
\frac{\sin t}{4\cos^{2}\left(\frac{t}{4}-\frac{\pi}{8}\right)}\,{\rm
d}t.\]
By using the further change of variable $u=\frac{t}{4}-\frac{\pi}{8}$,
or otherwise, show that
\[I=4\surd{2}-\pi-2.\]
\noindent[You may assume that $\tan\frac{\pi}{8}=\surd{2}-1$.]

Solution source

\begin{align*}
&& \cos 4u &= 2\cos^2 2u - 1 \\
&&&= 2 (2\cos^2 u - 1)^2 - 1 \\
&&&= 2(4\cos^4u - 4\cos^2 u + 1) - 1\\
&&&= 8\cos^4u - 8\cos^2 u + 1 
\end{align*}

\begin{align*}
&& I &= \int_{-1}^1 \frac{1}{\sqrt{1+x}+\sqrt{1-x}+2} \d x \\
x = \cos t, \d x = - \sin t \d t: &&&= \int_{t = \pi}^{t=0} \frac{1}{\sqrt{1+\cos t} + \sqrt{1-\cos t} + 2} (- \sin t ) \d t \\
&&&= \int_0^\pi \frac{\sin t}{\sqrt{2 \cos^2 \frac{t}{2}}+\sqrt{2 \sin^2 \frac{t}{2}}+2} \d t \\
&&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\cos \frac{t}{2} + \sin \frac{t}{2}) + 2} \d t \\
&&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\sqrt{2} \cos (\frac{t}{2}-\frac{\pi}{4})) + 2} \d t \\
&&&= \int_0^\pi \frac{\sin t}{2(1+\cos (\frac{t}{2}-\frac{\pi}{4}))} \d t \\
&&&= \int_0^\pi \frac{\sin t}{4\cos^2(\frac{t}{4}-\frac{\pi}{8})} \d t \\
\\
u = \tfrac{t}{4} -\tfrac{\pi}{8}, \d u = \tfrac14 \d t:&&&=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin (4u+\frac{\pi}{2})}{4 \cos^2 u} 4 \d u \\
&&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\cos4u}{\cos^2 u}  \d u \\
&&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 (2 \cos^2 u-1)-4 + \sec^2 u  \d u \\
&&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 \cos 2u-4 + \sec^2 u  \d u \\
&&&= \left [2\sin 2u - 4u + \tan u \right]_{-\pi/8}^{\pi/8} \\
&&&= 4 \sin \frac{\pi}{4} - \pi+ 2\tan \frac{\pi}{8} \\
&&&= \frac{4}{\sqrt{2}} - \pi + 2\sqrt{2}-2 \\
&&&= 4\sqrt{2} - \pi - 2
\end{align*}