Subject: Mathematics — Definite Integration
Question: Prove that for any positive integer $k$,
$\frac{sin2kx}{sinx} = 2\lbrack cosx + cos3x + \ldots\ldots\ldots\ldots.. + cos(2k - 1)x\rbrack$
Evaluate $\int_{0}^{\pi/2}\mspace{2mu}\sin{2kx}cotx\ dx$
Answer: $\frac{\mathbf{\pi}}{\mathbf{2}}$
Solution: $\because 2sinx\lbrack cosx + cos3x + cos5x\ldots + cos(2k - 1)x\rbrack$\
$${= 2sinxcosx + 2sinxcos3x + 2sinxcos5x
}{+ \ldots + 2sinxcos(2k - 1)x
}{= sin2x + (sin4x - sin2x)\ + (sin6x - sin4x)
}{+ \ldots + \{ sin2kx - sin(2k - 2)x\}
}{= sin2kx
}{\therefore 2\lbrack cosx + cos3x + cos5x + \ldots + cos(2k - 1)x\rbrack
}{= \frac{sin2kx}{sinx}\ldots.(i)}$$
Now, $sin2kx \cdot cotx = \frac{sin2kx}{sinx} \cdot cosx$\
$= 2cosx\lbrack cosx + cos3x + cos5x + \ldots + cos(2k - 1)x\rbrack$
\[from eqn (i)\]\
$${= \left\lbrack 2\cos^{2}x + 2cosxcos3x + 2cosxcos5x + \right.\ 
}{= (1 + cos2x) + (cos4x + cos2x)\ldots + 2cosxcos(2k - 1)x\rbrack
}{+ (cos6x + cos4x) + \ldots + \{ cos2kx + cos(2k - 2)x\} = 1 + 2\lbrack cos2x + cos4x + cos6x + \ldots + cos(2k - 2)x\rbrack + cos
}{\therefore\int_{0}^{\pi/2}\mspace{2mu}(sin2kx) \cdot cotxdx
}{= \int_{0}^{x/2}\mspace{2mu} 1 \cdot dx + 2\int_{0}^{\pi/2}\mspace{2mu}(cos2x + cos4x\ldots cos(2k - 2)x)dx + \int_{0}^{\pi/2}\mspace{2mu} cos(2k)xdx}$$
$= \frac{\pi}{2} + 2\left\lbrack \frac{\sin{2x}}{2} + \frac{\sin{4x}}{4} + \ldots + \frac{\sin(2k - 2)x}{(2k - 2)} \right\rbrack_{0}^{\frac{\pi}{2}}$\
$$+ \left\lbrack \frac{\sin(2k)x}{2k} \right\rbrack_{0}^{\frac{\pi}{2}}$$
$$= \frac{\pi}{2} + 2 \times 0 = \frac{\pi}{2}$$