\(% 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}}\)
Using integration by parts, we have \begin {align*} I(a, b) & = \int _{0}^{\frac {\pi }{2}} \cos ^a x \cos bx \Diff x \\ & = \frac {1}{b} \int _{0}^{\frac {\pi }{2}} \cos ^a x \Diff \sin bx \\ & = \frac {1}{b} \left [\left (\cos ^a x \sin bx\right )_{0}^{\frac {\pi }{2}} - \int _{0}^{\frac {\pi }{2}} \sin bx \Diff \cos ^a x \right ] \\ & = - \frac {1}{b} \int _{0}^{\frac {\pi }{2}} \sin bx \Diff \cos ^a x \\ & = \frac {a}{b} \int _{0}^{\frac {\pi }{2}} \sin bx \sin x \cos ^{a - 1} x \Diff x. \end {align*}
Notice that \[ \cos (b - 1) x = \cos bx \cos x + \sin bx \sin x, \] and hence \begin {align*} I(a - 1, b - 1) & = \int _{0}^{\frac {\pi }{2}} \cos ^{a - 1} x \cos (b - 1)x \Diff x \\ & = \int _{0}^{\frac {\pi }{2}} \cos ^{a - 1} x (\cos bx \cos x + \sin bx \sin x) \Diff x \\ & = \int _{0}^{\frac {\pi }{2}} \cos ^a x \cos bx \Diff x + \int _{0}^{\frac {\pi }{2}} \sin bx \sin x \cos ^{a - 1} x \Diff x \\ & = I(a, b) + \frac {b}{a} I(a, b) \\ & = \frac {a + b}{a} I(a, b), \end {align*}
and hence \[ I(a, b) = \frac {a}{a + b} I(a - 1, b - 1), \] as desired.
We look at the base case where \(n = 0\), and we have \begin {align*} \LHS & = \int _{0}^{\frac {\pi }{2}} \cos (2m + 1)x \Diff x \\ & = \frac {1}{2m + 1} \left [\sin (2m + 1)x\right ]_{0}^{\frac {\pi }{2}} \\ & = \frac {1}{2m + 1} \sin \frac {(2m + 1)\pi }{2} \\ & = \frac {(-1)^m}{2m + 1}, \end {align*}
and \[ \RHS = (-1)^m \frac {2^0 0! (2m)! m!}{m!(2m + 1)!} = \frac {(-1)^m}{2m + 1}, \] and so \(\LHS = \RHS \), which means this holds for the base case where \(n = 0\).
Now assume this is true for some \(n = k \geq 0\), i.e. \[ I(k, 2m + k + 1) = (-1)^m \frac {2^k k! (2m)! (k + m)!}{m! (2k + 2m + 1)!}, \] and we look at the case \(n = k + 1\). Note that \begin {align*} \LHS & = I(k + 1, 2m + k + 2) \\ & = \frac {k + 1}{2m + 2k + 3} I(k, 2m + k + 1) \\ & = \frac {k + 1}{2m + 2k + 3} (-1)^m \frac {2^k k! (2m)! (k + m)!}{m! (2k + 2m + 1)!} \\ & = (-1)^m \frac {2^k k! (2m)! (k + m)! (k + 1)}{m! (2k + 2m + 1)! (2k + 2m + 3)} \\ & = (-1)^m \frac {2^k (k + 1)! (2m)! (k + m)! 2 (k + m + 1) }{m! (2k + 2m + 1)! (2k + 2m + 2) (2k + 2m + 3)} \\ & = (-1)^m \frac {2^{k + 1} (k + 1)! (2m)! (k + m + 1)!}{m! (2k + 2m + 3)!} \\ & = (-1)^m \frac {2^{k + 1} (k + 1)! (2m)! [(k + 1) + m]!}{m! (2(k + 1) + 2m + 1)!}, \end {align*}
which shows the original statement is true for \(n = k + 1\).
Hence, by the principle of mathematical induction, the original statement is true for any non-negative integers \(n, m\).