How about Chebyshev?

Calculus Level 3

Compute

∫ 0 π / 2 sin ⁡ ( 2015 x ) sin ⁡ ( x ) d x . \large \int_0^{\pi /2} \frac{ \sin(2015x) }{\sin(x)} \, dx.

π 3 \frac\pi3 π 4 \frac\pi4 π 2 \frac\pi2 π 6 \frac\pi6

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Otto Bretscher
Nov 14, 2015

∫ sin ⁡ ( ( 2 n + 1 ) x ) sin ⁡ x d x = x + ∑ k = 1 n 1 k sin ⁡ ( 2 k ) + C \int \frac{\sin((2n+1)x)}{\sin{x}}dx=x+\sum_{k=1}^{n}\frac{1}{k}\sin(2k)+C

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah. To be clear, Mr (Professor) Otto Bretscher used the identity

2 sin ⁡ ( x ) [ 1 + cos ⁡ ( 2 x ) + cos ⁡ ( 4 x ) + … + cos ⁡ ( 2 n x ) ] = sin ⁡ [ ( 2 n + 1 ) x ] + sin ⁡ ( x ) . 2\sin (x) \left [ 1 + \cos(2x) + \cos(4x) + \ldots + \cos(2nx) \right] = \sin[(2n+1)x] + \sin(x).

Alternatively, one can prove this by applying sin ⁡ ( x ) = 1 2 ( e i x − e − i x ) \sin(x) = \frac12 (e^{ix} - e^{-ix} ) .

Pi Han Goh - 5 years, 7 months ago

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I was integrating the Dirichlet Kernel sin ⁡ ( ( 2 n + 1 ) x ) sin ⁡ x = ∑ k = − n n e 2 k i x = 1 + 2 ∑ k = 1 n cos ⁡ ( 2 k x ) \frac{\sin((2n+1)x)}{\sin x}=\sum_{k=-n}^{n}e^{2k ix}=1+2\sum_{k=1}^{n}\cos(2kx)

Please stop this ridiculous "Mr (Prof) O B " business and call me "Comrade Otto" ;)

Otto Bretscher - 5 years, 7 months ago

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Affirmative, Comrade Otto.

Pi Han Goh - 5 years, 7 months ago

WİRKLİCH COMRADE

Soner Karaca - 5 years, 6 months ago

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