Evaluate $\displaystyle \int_0^{\pi/2}\sin (2nx)\cot x dx$
$$\sin (2nx)\cot x=\frac{\sin (2nx)}{\sin x}\cos x$$
I know that
$$\frac{\sin (\frac{k\beta}{2})}{\sin (\frac{\beta}{2})}\cos\left(\alpha+\frac{k-1}{2}\beta\right)=\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+(k-1)\beta)$$
So comparing between $\frac{\sin (2nx)}{\sin x}\cos x$ and $\frac{\sin (\frac{k\beta}{2})}{\sin (\frac{\beta}{2})}\cos\left(\alpha+\frac{k-1}{2}\beta\right)$ I got $\beta=2x$, $k=2n$ and $\alpha=-(2n-2)x$.
Therefore,
$$\begin{align*}\frac{\sin (2nx)}{\sin x}\cos x&=\cos(-[2n-2]x)+\cos(-[2n-2]x+2x)+\cos(-[2n-2]x+4x)+\cdots+\cos(2nx)\\&=\cos0+\cos(2nx)+2[\cos(2x)+\cos(4x)+\cdots+\cos((2n-2)x)]\\&=1+\cos(2nx)+2[\cos(2x)+\cos(4x)+\cdots+\cos((2n-2)x)]\end{align*}$$
$$\int_0^{\pi/2}\cos 2kx dx=0$$
$$\implies\int_0^{\pi/2}\sin (2nx)\cot x dx=\int_0^{\pi/2}dx=\frac{\pi}{2}$$
Can someone provide an alternative solution without using summation of cosines. Thanks in advance.
Best Answer
Let $$f_n=\int_{0}^{\frac \pi 2} \sin (2nx)\cot (x)\;\text{d}x$$ We will first show that $f_{n+1}-f_n=0$ for any $n\in \mathbb N$.
\begin{align*} f_{n+1}-f_n&=\int_{0}^{\frac \pi 2}(\sin(2(n+1)x)-\sin(2nx))\cot(x)\;\text{d}x\\ &=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\sin\left(\frac{2(n+1)x-2nx}2\right)\cot(x)\;\text{d}x\\ &=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\sin(x)\frac{\cos (x)}{\sin (x)}\;\text{d}x\\ &=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\cos (x)\;\text{d}x\\ &=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\cos\left(\frac{2(n+1)x-2nx}2\right)\;\text{d}x\\ &=\int_{0}^{\frac \pi 2} \cos(2(n+1)x)+\cos(2nx))\;\text{d}x\\ &=\frac 1{2(n+1)}\sin(2(n+1)x)+\frac 1{2n}\sin(2nx)\bigg|_{0}^{\frac \pi 2}\\ &=\frac 1{2(n+1)}\sin((n+1)\pi)+\frac 1{2n}\sin(n\pi)-\frac 1{2(n+1)}\sin(0)-\frac 1{2n}\sin(0)\\ &=0 \end{align*}
Now, just calculate $f_1=\frac \pi 2$ and complete the proof