Evaluate the integral $\int_0^{\pi/2} \sin (2n x) \tan x\, dx$
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int_{0}^{\pi/2}\sin\pars{2nx}\tan\pars{x}\,\dd x} =\sgn\pars{n}\,\Im\int_{0}^{\pi/2}\bracks{\expo{2\verts{n}x\ic} - \pars{-1}^{\verts{n}}}\tan\pars{x}\,\dd x \\[5mm]&=\sgn\pars{n}\,\Im \int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ \,{\rm Arg}\pars{z}\ <\ \pi/2}} \bracks{z^{2\verts{n}} - \pars{-1}^{\verts{n}}} {\pars{z^{2} - 1}/\pars{2\ic z} \over \pars{z^{2} + 1}/\pars{2z}} \,{\dd z \over \ic z} \\[5mm]&=\sgn\pars{n}\,\Im \int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ \,{\rm Arg}\pars{z}\ <\ \pi/2}} \bracks{z^{2\verts{n}} - \pars{-1}^{\verts{n}}} {1 - z^{2} \over 1 + z^{2}}\,{\dd z \over z} \\[5mm]&=-\sgn\pars{n}\left.\lim_{\epsilon\ \to\ 0^{+}}\ \int_{\pi/2}^{0} \bracks{z^{2\verts{n}} - \pars{-1}^{\verts{n}}} {1 - z^{2} \over 1 + z^{2}}\,{\dd z \over z} \right\vert_{\, z\ \equiv\ \epsilon\expo{\ic\theta}} =\sgn\pars{n}\pars{-1}^{\verts{n}}\,\Im\int_{\pi/2}^{0}\ic\,\dd\theta \\[5mm]&=\sgn\pars{n}\pars{-1}^{\verts{n} + 1}\,\,{\pi \over 2} =\color{#66f}{\large\sgn\pars{n}\pars{-1}^{n + 1}\,{\pi \over 2}}\,,\qquad n \in {\mathbb Z}. \end{align}
There are two integrals $\ds{\pars{~\mbox{from}\ y=1\ \mbox{to}\ y=\epsilon\ \mbox{and from}\ x=\epsilon\ \mbox{to}\ 1~}}$ which don't yield any contribution because they are reals.
The integration was performed by 'closing' a contour in the first quadrant.
'Real' Method:
With $\ds{\left.I_{n}\,\right\vert_{\, n\ \geq\ 1} \equiv \half\int_{-\pi/2}^{\pi/2}\sin\pars{2nx}\tan\pars{x}\,\dd x}$ we'll have: \begin{align} I_{n + 1} - I_{n - 1}&= \int_{-\pi/2}^{\pi/2}\cos\pars{2nx}\sin\pars{2x}\tan\pars{x}\,\dd x =2\int_{-\pi/2}^{\pi/2}\cos\pars{2nx}\sin^{2}\pars{x}\,\dd x \\[5mm]&=\int_{-\pi/2}^{\pi/2}\bracks{\cos\pars{2nx} - \cos\pars{2nx}\cos\pars{2x}}\,\dd x =0 \end{align} Then $\ds{I_{2n\ \geq\ 2}\,\,\, =\ I_{2}}$ and $\ds{I_{2n + 1\ \geq\ 1}\,\,\,=\ I_{1}}$
Hint. Here is an approach.
Step 1. Observe that, for $x \in (-\pi/2,\pi/2)$, we have $$ \begin{align} \log (\cos x)&=\log \left(\frac{e^{ix}+e^{-ix}}{2}\right)\\\\ &=-\log 2+\Re\log \left(1+e^{2ix}\right)\\\\ &=-\log 2+\Re \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} e^{2inx}\\\\ &=-\log 2+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\cos(2nx) \tag1 \end{align} $$
Step 2. Performing an integration by parts $$ \int_0^{\pi/2} \sin (2n x) \tan x \: dx=\left.-\sin (2n x) \log (\cos x) \right|_0^{\pi/2}+2n\int_0^{\pi/2} \cos (2n x) \log (\cos x) \: dx $$ giving $$ \int_0^{\pi/2} \sin (2n x) \tan x \: dx=2n\int_0^{\pi/2} \!\! \cos (2n x) \log (\cos x) \: dx \tag2 $$
Step 3. By uniqueness of the Fourier coefficients, using $(1)$ and $(2)$, you just get
$$ \int_0^{\pi/2} \sin (2n x) \tan x \: dx=(-1)^{n-1}\frac{\pi}{2}, \qquad n=1,2,3,\ldots \tag3 $$
Using the trigonometric identities $$ \sin(x)+\sin(y)=2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y\vphantom{+}}2\right)\tag{1} $$ and $$ 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y)\tag{2} $$ we get $$ \begin{align} [\sin(2(n+1)x)+\sin(2nx)]\tan(x) &=2\sin((2n+1)x)\cos(x)\tan(x)\\ &=2\sin((2n+1)x)\sin(x)\\ &=\cos(2nx)-\cos(2(n+1)x)\tag{3} \end{align} $$ For $n\ge1$, we have $$ \int_0^{\pi/2}\cos(2nx)\,\mathrm{d}x=0\tag{4} $$
Therefore, integrating $(3)$ and applying $(4)$ yields $$ \int_0^{\pi/2}\sin(2(n+1)x)\tan(x)\,\mathrm{d}x =-\int_0^{\pi/2}\sin(2nx)\tan(x)\,\mathrm{d}x\tag{5} $$ then noting $$ \begin{align} \int_0^{\pi/2}\sin(2x)\tan(x)\,\mathrm{d}x &=2\int_0^{\pi/2}\sin(x)\cos(x)\tan(x)\,\mathrm{d}x\\ &=2\int_0^{\pi/2}\sin^2(x)\,\mathrm{d}x\\ &=\int_0^{\pi/2}(1-\cos(2x))\,\mathrm{d}x\\[4pt] &=\frac\pi2\tag{6} \end{align} $$ gives that for $n\ge1$, $$ \int_0^{\pi/2}\sin(2nx)\tan(x)\,\mathrm{d}x=(-1)^{n-1}\frac\pi2\tag{7} $$