How may we show that $\int_{0}^{\pi/2}{\mathrm dx\over (1+\tan x)^6}={11\over 6}\cdot{1\over 10}?$ proposed:

$$\int_{0}^{\pi/2}{\mathrm dx\over (1+\tan x)^6}={11\over 6}\cdot{1\over 10}\tag1$$

My try:
$$\int_{0}^{\pi/2}{\cos^6(x)\over (\sin(x)+\cos(x))^6}\tag2$$
$u=\cos(x)\implies du=-6\cos^5(x)\sin(x)\ dx$
$${1\over 6}\int_{0}^{1}{u\over u^{1/6}+\sqrt{1-u^{1/3}}}\cdot{\mathrm du\over u^{5/6}\sqrt{1-u^{1/3}}}\tag3$$
Getting too complicated.
How can we prove $(1)?$
 A: The interval should be $[0,\pi/2]$ instead of $[0,\pi]$. Splitting $[0,\pi/2]$ into $[0,\pi/4]\cup[\pi/4,\pi/2]$, one has
\begin{eqnarray}
&&\int_{0}^{\pi/2}{\mathrm dx\over (1+\tan x)^6}\\
&=&\int_{0}^{\pi/4}{\mathrm dx\over (1+\tan x)^6}+\int_{\pi/4}^{\pi/2}{\mathrm dx\over (1+\tan x)^6}\\
&=&\int_{0}^{\pi/4}{\mathrm dx\over (1+\tan x)^6}+\int_0^{\pi/4}{\mathrm dx\over (1+\cot x)^6}\\
&=&\int_{0}^{\pi/4}{1+\tan^6x\over (1+\tan x)^6}\mathrm dx\\
&=&\int_0^1\frac{1+u^6}{(1+u^2)(1+u)^6}\mathrm dx\\
&=&\int_0^1\frac{1-u^2+u^4}{(1+u)^6}\mathrm dx\\
&=&\int_1^2\frac{1-(u-1)^2+(u-1)^4}{u^6}\mathrm dx\\
&=&\int_1^2\bigg(\frac{1}{u^6}-\frac{2}{u^5}+\frac{5}{u^4}-\frac{4}{u^3}+\frac{1}{u^2}\bigg)du\\
&=&\frac{11}{60}.
\end{eqnarray}
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[15px,#ffe]{\ds{%
\int_{0}^{\pi/2}{\dd x \over \bracks{1 + \tan\pars{x}}^{\,6}}}} =
\int_{-\pi/4}^{\pi/4}{\dd x \over \bracks{1 + \tan\pars{x + \pi/4}}^{\,6}}
\\[5mm] = &\
\int_{-\pi/4}^{\pi/4}{\dd x \over
\braces{1 + \bracks{\tan\pars{x} + 1}/\bracks{1 - \tan\pars{x}}}^{\,6}} =
{1 \over 64}\int_{-\pi/4}^{\pi/4}\bracks{1 - \tan\pars{x}}^{\,6}\,\dd x
\\[5mm] = &\
{1 \over 64}\int_{0}^{\pi/4}\braces{\bracks{1 - \tan\pars{x}}^{\,6} +
\bracks{1 + \tan\pars{x}}^{\,6}}\,\dd x =
{1 \over 64}\pars{2I_{0} + 30I_{2} + 30I_{4} + 2I_{6}}
\\[5mm] = &
{1 \over 32}\pars{I_{0} + 15I_{2} + 15I_{4} + I_{6}}
\end{align}

where
  $\ds{I_{n}  \equiv \int_{0}^{\pi/4}\tan^{n}\pars{x}\,\dd x =
\int_{0}^{\pi/4}\tan^{n - 2}\pars{x}\sec^{2}\pars{x}\,\dd x -
\int_{0}^{\pi/4}\tan^{n - 2}\pars{x}\,\dd x = {1 \over n - 1} - I_{n - 2}}$.


$$
\left\{\begin{array}{rcccl}
\ds{I_{0}} & \ds{=} & \ds{\pi \over 4} &&
\\[2mm]
\ds{I_{2}} & \ds{=} & \ds{1 - I_{0}} & \ds{=} &
\ds{\phantom{-}1 - {\pi \over 4}} 
\\[2mm]
\ds{I_{4}} & \ds{=} & \ds{{1 \over 3} - I_{2}} & \ds{=} &
\ds{-\,{2 \over 3} + {\pi \over 4}}
\\[2mm]
\ds{I_{6}} & \ds{=} & \ds{{1 \over 5} - I_{4}} & \ds{=} &
\ds{\phantom{-}{13 \over 15} - {\pi \over 4}} 
\end{array}\right.
$$

\begin{align}
\bbox[15px,#ffe]{\ds{%
\int_{0}^{\pi/2}{\dd x \over \bracks{1 + \tan\pars{x}}^{\,6}}}} & =
{1 \over 32}\bracks{%
{\pi \over 4} + 15\pars{1 - {\pi \over 4}} + 15\pars{-\,{2 \over 3} +
{\pi \over 4}} + \pars{{13 \over 15} - {\pi \over 4}}}
\\[5mm] & =
{1 \over 32}\pars{15 - 10 + {13 \over 15}} = \bbx{11 \over 60}
\end{align}
A: Hint:
To compute the antiderivative, Bioche's rules say we have to set $t=\tan x$. Indeed
$$\mathrm d t=(1+t^2)\,\mathrm d x,\enspace\text{so}\quad \int\frac{\mathrm dx}{ (1+\tan x)^6}=\int\frac{\mathrm dx}{(1+t^2) (1+t)^6}.$$
To obtain the integral of this rational function, obtain first the partial fractions decomposition
$$\frac1{(1+t^2) (1+t)^6}=\frac{At+B}{1+t^2}+\sum_{i=1}^6\frac{C_i}{(1+t)^i}.$$
This is most easily obtained  setting $u=1+t$, so that
$$\frac1{(1+t^2) (1+t)^6}=\frac1{(2-2u+u^2) u^6}=\frac{A'u+B'}{2-2u+u^2}+\sum_{i=1}^6\frac{C'_i}{u^i},$$
and performing the division of $1$ by $2-2u+u^2$ by increasing powers up to order $6$, so that
$$1=Q(u)(2-2u+u^2)+u^6R(u)\qquad(\deg Q(u)\le 5)$$
and finally
$$\frac1{(2-2u+u^2) u^6}=\frac{Q(u)}{u^6}+\frac{R(u)}{2-2u+u^2}.$$
