Trigonometry! #139

Let $\displaystyle I(k) = \int_0^{\pi /2} \sin(2kx) \cot x \, dx$ . We will attempt to prove that $I(k) = \dfrac{\pi}2$ for $k=1,2,3,\ldots$ .

Lemma 1: $I(1) = \dfrac{\pi}2$ .

Proof: $I(1) = \int_0^{\pi/2} \sin (2x) \cot x \, dx =\int_0^{\pi/2} 2\sin x \cos \cdot \dfrac{\cos x}{\sin x} \, dx = \int_0^{\pi/2} 2\cos^2 x \, dx = \int_0^{\pi/2} (\cos(2x) + 1) \, dx = \dfrac{\pi}2 \qquad \blacksquare$

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Lemma 2: $I(k+1) - I(k) = 0$ for all $k=1,2,3,\ldots$ .

Proof: For the equation below, we use apply the two trigonometric identities: $\sin(A) - \sin(B) = 2\cos\left(\dfrac{A+B}2 \right) \sin \left(\dfrac{A-B}2 \right)$ and $2\cos A \cos B = \cos(A+B) + \cos(A-B)$ .

$\begin{aligned} I(k + 1) - I(k) &=& \int_0^{\pi /2} \left [ \sin((2k+1)x) - \sin(2kx) \right ] \cot x \, dx \\ &=& \int_0^{\pi /2} \left [ 2\cos((k+1)x) \cdot \sin x \right ] \cdot \dfrac{\cos x}{\sin x}\, dx \\ &=& \int_0^{\pi /2} 2\cos((k+1)x) \cos x \, dx \\ &=& \int_0^{\pi /2} \left [ \cos((2k+2)x) + \cos(2kx) \right ] \, dx = 0 \qquad \blacksquare \end{aligned}$

From Lemma 2, we know that $0 = I(2) - I(1) = I(3) - I(2) = I(4) - I(3) =\cdots$ , thus $I(k) = \dfrac{\pi}2$ . The answer is $I(k) \div \pi = \boxed{0.5}$ .