Problem source

The integral $I$ is defined by 
\[
I=\int_{1}^{2}\frac{(2-2x+x^{2})^{k}}{x^{k+1}}\,\mathrm{d}x
\]
where $k$ is a constant. Show that
\[
I=\int_{0}^{1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x=\int_{0}^{\frac{1}{4}\pi}\frac{\mathrm{d}\theta}{\left[\sqrt{2}\cos\theta\cos\left(\frac{1}{4}\pi-\theta\right)\right]^{k+1}}=2\int_{0}^{\frac{1}{8}\pi}\frac{\mathrm{d}\theta}{\left[\sqrt{2}\cos\theta\cos\left(\frac{1}{4}\pi-\theta\right)\right]^{k+1}}.
\]
Hence show that 
\[
I=2\int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x
\]
Deduce that
\[
\int_{1}^{\sqrt{2}}\left(\frac{2-2x^{2}+x^{4}}{x^{2}}\right)^{k}\frac{1}{x}\,\mathrm{d}x=\int_{1}^{\sqrt{2}}\left(\frac{2-2x+x^{2}}{x}\right)^{k}\frac{1}{x}\,\mathrm{d}x
\]

Solution source

\begin{align*}
I &=\int_{1}^{2}\frac{(2-2x+x^{2})^{k}}{x^{k+1}}\,\mathrm{d}x \\
u = x-1 &, \quad \d u = \d x \\
&= \int_{u = 0}^{u=1} \frac{(u^2+1)^k}{(u+1)^{k+1}} \d u \\
&= \boxed{\int_0^1 \frac{(1+x^2)^k}{(1+x)^{k+1}} \d x} \\
x = \tan \theta &, \quad \d x = \sec^2 \theta \d \theta \\
&= \int_{\theta = 0}^{\theta = \pi/4} \frac{\sec^{2k+2} \theta }{(1 + \tan \theta)^{k+1}} \d \theta \\
&= \int_0^{\pi/4} \frac{\d \theta}{\cos^{2k+2} \theta (\frac{\sin \theta + \cos \theta}{\cos \theta})^{k+1}} \\
&= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta ({\sin \theta + \cos \theta})^{k+1}} \\
&= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta (\sqrt{2} \cos (\frac{\pi}{4} - \theta))^{k+1}} \\
I &= \boxed{ \int_0^{\pi/4} \frac{\d \theta}{(\sqrt{2}\cos \theta  \cos (\frac{\pi}{4} - \theta))^{k+1}}} \\
\end{align*}

Since $f(\theta) = \cos \theta \cos (\frac{\pi}{4} - \theta)$ is symmetric about $\frac{\pi}{8}$ this integral is twice the integral to $\frac{\pi}{8}$.

$\tan 2 \theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \Rightarrow 1 = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}} \Rightarrow \tan \frac{\pi}{8} = \sqrt{2}-1$.

Therefore, using the same substitution we must have:

\[
I=2\int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x
\]

Let $u = x^2$, then $\d u = 2 x\d x$

\begin{align*}
\int_{1}^{\sqrt{2}}\left(\frac{2-2x^{2}+x^{4}}{x^{2}}\right)^{k}\frac{1}{x}\,\mathrm{d}x &= \int_{u = 1}^{u = 2} \l \frac{2-2u+u^2}{u}\r^k \frac{1}{2u} \d u \\
&= \frac12 I \\
&= \int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \\
u = 1+x & \quad \d u = \d x \\
&= \int_1^{\sqrt{2}} \frac{(1+(u-1)^2)^k}{u^{k+1}} \d u \\
&= \int_{1}^{\sqrt{2}}\left(\frac{2-2u+u^{2}}{u}\right)^{k}\frac{1}{u}\,\mathrm{d}x \\
&= \int_{1}^{\sqrt{2}}\left(\frac{2-2x+x^{2}}{x}\right)^{k}\frac{1}{x}\,\mathrm{d}x
\end{align*}