Calculus Generalisation

Let $$J(n) = \int_0^{\pi/2} \cos^n(x) \sin((n+2) x)\; dx $$ Using complex exponentials, we can write this as $ J(n) = \text{Im}(A(n))$, where $$ \eqalign{ A(n) &= \int_0^{\pi/2} \cos^n(x) e^{i(n+2)x} \; dx \cr &= \int_0^{\pi/2} e^{2ix} 2^{-n}\left( e^{2ix+1}+1\right)^n\; dx}$$ The generating function of this, which should converge at least for $|t|<1$, is $$ \eqalign{f(t) &= \sum_{n=0}^\infty A(n) t^n \cr &= \int_0^{\pi/2} e^{2ix} \sum_{n=0}^\infty ((e^{2ix}+1) t/2)^n \; dx\cr &= \int_0^{\pi/2} \frac{2 e^{2ix}}{2-t-t e^{2ix}} \; dx\cr & = \left. \frac{i \ln(2-t-t e^{2ix})}{t}\right|_{x=0}^{\pi/2}\cr &= \frac{-i \ln(1-t)}{t}\cr &= i \sum_{n=0}^\infty \frac{t^n}{n+1}}$$

Thus $A(n) = i/(n+1)$ and $J(n) = 1/(n+1)$.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\pi/2}\cos^{2011}\pars{x}\sin\pars{2013x}\,\dd x = \Im\int_{0}^{\pi/2}\cos^{2011}\pars{x}\expo{2013x\ic}\,\dd x \\[5mm] = &\ \left.\Im\int_{x = 0}^{x = \pi/2}\pars{z + 1/z \over 2}^{2011}z^{2013} \,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.-\,{1 \over 2^{2011}}\,\Re\int_{x = 0}^{x = \pi/2}\pars{1 + z^{2}}^{2011} \,z\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ {1 \over 2^{2011}}\,\Re\int_{1}^{0}\pars{1 - y^{2}}^{2011} \pars{\ic y}\ic\,\dd y + \,{1 \over 2^{2011}}\,\Re\int_{0}^{1}\pars{1 + x^{2}}^{2011}\ x\,\dd x \\[5mm] = &\ {1 \over 2^{2012}}\int_{0}^{1}\pars{1 - y}^{2011}\,\dd y + \,{1 \over 2^{2012}}\int_{0}^{1}\pars{1 + x}^{2011}\,\dd x \\[5mm] = &\ {1 \over 2^{2012}}\,{1 \over 2012} + {1 \over 2^{2012}}\,{2^{2012} - 1 \over 2012} = \bbx{1 \over 2012} \end{align}