Indefinite Integration 1 Question 78
79. Prove that for any positive integer $k$,
$\frac{\sin 2 k x}{\sin x}=2[\cos x+\cos 3 x+\ldots+\cos (2 k-1) x]$
Hence, prove that $\int _0^{\pi / 2} \sin 2 k x \cdot \cot x d x=\pi / 2$.
$(1990,4$ M)
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Solution:
- We know that,
$2 \sin x[\cos x+\cos 3 x+\cos 5 x+\ldots+\cos (2 k-1) x]$
$=2 \sin x \cos x+2 \sin x \cos 3 x+2 \sin x \cos 5 x$
$+\ldots+2 \sin x \cos (2 k-1) x$
$=\sin 2 x+(\sin 4 x-\sin 2 x)+(\sin 6 x-\sin 4 x)$
$+\ldots+{\sin 2 k x-\sin (2 k-2) x}$
$=\sin 2 k x$
$\therefore \quad 2[\cos x+\cos 3 x+\cos 5 x+\ldots+\cos (2 k-1) x]$
$=\frac{\sin 2 k x}{\sin x}$
Now, $\quad \sin 2 k x \cdot \cot x=\frac{\sin 2 k x}{\sin x} \cdot \cos x$
$=2 \cos x[\cos x+\cos 3 x+\cos 5 x+\ldots+\cos (2 k-1) x]$
[from Eq. (i)]
$=\left[2 \cos ^{2} x+2 \cos x \cos 3 x+2 \cos x \cos 5 x+\right.$ . $+2 \cos x \cos (2 k-1) x]$
$=(1+\cos 2 x)+(\cos 4 x+\cos 2 x)$
$+(\cos 6 x+\cos 4 x)+\ldots+{\cos 2 k x+\cos (2 k-2) x}$
$=1+2[\cos 2 x+\cos 4 x+\cos 6 x+\ldots+\cos (2 k-2) x]$
$+\cos 2 k x$
$\therefore \int _0^{\pi / 2}(\sin 2 k x) \cdot \cot x d x$
$=\int _0^{\pi / 2} 1 \cdot d x+2 \int _0^{\pi / 2}(\cos 2 x+\cos 4 x \ldots \cos (2 k-2) x) d x$ $+\int _0^{\pi / 2} \cos (2 k) x d x$
$=\frac{\pi}{2}+2 \frac{\sin 2 x}{2}+\frac{\sin 4 x}{4}+\ldots+\frac{\sin (2 k-2) x}{(2 k-2)} _0^{\pi / 2}$
$+\frac{\sin (2 k) x}{2 k}{ } _0^{\pi / 2}=\frac{\pi}{2}$
