\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since \(t = \tan \frac {1}{2}x\), we have \[ \DiffFrac {t}{x} = \frac {1}{2} \sec ^2 \frac {1}{2}x = \frac {1}{2} (1 + \tan ^2 \frac {1}{2}x) = \frac {1}{2}(1 + t^2). \]
By the tangent double-angle formula, we have \[ \tan x = \frac {2t}{1 - t^2}, \] and hence \[ \cot x = \frac {1 - t^2}{2t}. \]
Therefore, \[ \csc ^2 x = 1 + \cot ^2 x = 1 + \frac {(1 - t^2)^2}{(2t)^2} = \frac {(1 + t^2)^2}{(2t)^2}, \] which means \[ \sin ^2 x = \frac {(2t)^2}{(1 + t^2)^2}, \] and hence \[ \abs *{\sin x} = \frac {2t}{1 + t^2}. \]
What remains is to consider the sign. Notice that \(t \geq 0\) if and only if \[ \frac {x}{2} \in \bigcup _{k \in \ZZ } \left [k\pi , k\pi + \frac {\pi }{2}\right ), \] which is \[ x \in \bigcup _{k \in \ZZ } \left [2k\pi , 2k\pi + \pi \right ), \] but this is also precisely if and only if \(\sin x \geq 0\).
This means \(\sin x\) must take the same sign as \(t\), and hence \[ \sin x = \frac {2t}{1 + t^2}. \]
Using this substitution, we have when \(x = 0, t = 0\) and when \(x = \frac {1}{2}\pi , t = 1\), and also \[ \Diff x = \frac {2 \Diff t}{1 + t^2}. \]
This means \begin {align*} I & = \int _{0}^{\frac {1}{2}\pi } \frac {\Diff x}{1 + a \sin x} \\ & = \int _{0}^{1} \frac {\frac {2 \Diff t}{1 + t^2}}{1 + a \cdot \frac {2t}{1 + t^2}} \\ & = \int _{0}^{1} \frac {2 \Diff t}{1 + 2at + t^2} \\ & = \int _{0}^{1} \frac {2 \Diff t}{(t + a)^2 + (1 - a^2)} \\ & = \frac {2}{1 - a^2} \int _{0}^{1} \frac {\Diff t}{\left (\frac {t + a}{\sqrt {1 - a^2}}\right )^2 + 1} \\ & = \frac {2}{1 - a^2} \cdot \sqrt {1 - a^2} \cdot \left [\arctan \left (\frac {t + a}{\sqrt {1 - a^2}}\right )\right ]_{0}^{1} \\ & = \frac {2}{\sqrt {1 - a^2}} \cdot \left [\arctan \left (\frac {1 + a}{\sqrt {1 - a^2}}\right ) - \arctan \left (\frac {a}{\sqrt {1 - a^2}}\right )\right ]. \end {align*}
But notice that \begin {align*} \arctan \left (\frac {1 + a}{\sqrt {1 - a^2}}\right ) - \arctan \left (\frac {a}{\sqrt {1 - a^2}}\right ) & = \arctan \left (\frac {\frac {1 + a}{\sqrt {1 - a^2}} - \frac {a}{\sqrt {1 - a^2}}}{1 + \frac {1 + a}{\sqrt {1 - a^2}} \cdot \frac {a}{\sqrt {1 - a^2}}}\right ) \\ & = \arctan \left (\frac {\frac {1}{\sqrt {1 - a^2}}}{1 + \frac {a + a^2}{1 - a^2}}\right ) \\ & = \arctan \left (\frac {\sqrt {1 - a^2}}{(1 - a^2) + (a + a^2)}\right ) \\ & = \arctan \left (\frac {\sqrt {1 - a} \cdot \sqrt {1 + a}}{1 + a}\right ) \\ & = \arctan \left (\frac {\sqrt {1 - a}}{\sqrt {1 + a}}\right ), \end {align*}
and hence \[ I = \frac {2}{\sqrt {1 - a^2}} \arctan \left (\frac {\sqrt {1 - a}}{\sqrt {1 + a}}\right ), \] as desired.
We have \begin {align*} I_{n + 1} + 2 I_n & = \int _{0}^{\frac {1}{2}\pi } \frac {\sin ^{n + 1}x + 2 \sin ^n x}{2 + \sin x} \Diff x \\ & = \int _{0}^{\frac {1}{2}\pi } \sin ^n x\Diff x. \end {align*}
Therefore, we have \begin {align*} I_3 + 2 I_2 & = \int _{0}^{\frac {1}{2}\pi } \sin ^2 x \Diff x \\ & = \int _{0}^{\frac {1}{2}\pi } \frac {1 - \cos 2x}{2} \Diff x \\ & = \left [\frac {1}{2} \cdot x - \frac {1}{4} \sin 2x\right ]_{0}^{\frac {1}{2}\pi } \\ & = \left (\frac {1}{2} \cdot \frac {\pi }{2} - \frac {1}{4} \sin \pi \right ) - \left (\frac {1}{4} \sin 0 - \frac {1}{2} \cdot 0\right ) \\ & = \frac {\pi }{4}, \end {align*}
\begin {align*} I_2 + 2 I_1 & = \int _{0}^{\frac {1}{2}\pi } \sin x \Diff x \\ & = [- \cos x]_{0}^{\frac {1}{2}\pi } \\ & = \left (- \cos \frac {1}{2}\pi \right ) - (- \cos 0) \\ & = (0) - (-1) \\ & = 1, \end {align*}
and \begin {align*} I_1 + 2 I_0 & = \int _{0}^{\frac {1}{2}\pi } \sin ^0 x \Diff x \\ & = [x]_{0}^{\frac {1}{2}\pi } \\ & = \frac {1}{2}\pi . \end {align*}
Also, notice that \begin {align*} I_0 & = \int _{0}^{\frac {1}{2}\pi } \frac {\Diff x}{2 + \sin x} \\ & = \frac {1}{2} \int _{0}^{\frac {1}{2}\pi } \frac {\Diff x}{1 + \frac {1}{2} \sin x} \\ & = \frac {1}{2} \cdot \frac {2}{\sqrt {1 - \left (\frac {1}{2}\right )^2}} \cdot \arctan \frac {\sqrt {1 - \frac {1}{2}}}{\sqrt {1 + \frac {1}{2}}} \\ & = \frac {1}{2} \cdot \frac {4}{\sqrt {3}} \cdot \arctan \frac {1}{\sqrt {3}} \\ & = \frac {2}{\sqrt {3}} \cdot \frac {\pi }{6} \\ & = \frac {\pi }{3\sqrt {3}}. \end {align*}
Hence, \begin {align*} I_3 & = \frac {\pi }{4} - 2 I_2 \\ & = \frac {\pi }{4} - 2 \cdot \left (1 - 2 I_1\right ) \\ & = \frac {\pi }{4} - 2 + 4 I_1 \\ & = \frac {\pi }{4} - 2 + 4 \left (\frac {1}{2}\pi - 2 I_0\right ) \\ & = \frac {\pi }{4} - 2 + 2\pi - 8 I_0 \\ & = \frac {9\pi }{4} - 2 - \frac {8\pi }{3 \sqrt {3}} \\ & = \left (\frac {9}{4} - \frac {8}{3 \sqrt {3}}\right ) \pi - 2. \end {align*}