the beauty in integration

Level 1

∫ 0 1 1 x 2 − 2 x cos ⁡ θ + 1 d x = ? \large \int_0^1 \frac 1{x^2-2x\cos \theta + 1} dx = ?

π − θ sin ⁡ θ \frac {π - θ}{\sin θ} π − θ cos ⁡ θ \frac {π - θ}{\cos θ} 2 ( π − θ ) sin ⁡ θ \frac {2(π - θ)}{\sin θ} π − θ 2 sin ⁡ θ \frac {π - θ}{2 \sin θ}

Ahpa Tsum by Ahpa Tsum , 20, France

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2 solutions

Joseph Newton
May 4, 2018

∫ 0 1 1 x 2 − 2 x cos ⁡ θ + 1 d x = ∫ 0 1 1 ( x − cos ⁡ θ ) 2 + 1 − cos ⁡ 2 θ d x = ∫ 0 1 1 sin ⁡ 2 θ + ( x − cos ⁡ θ ) 2 d x = [ 1 sin ⁡ θ tan ⁡ − 1 ( x − cos ⁡ θ sin ⁡ θ ) ] 0 1 using ∫ 1 a 2 + u 2 d u = 1 a tan ⁡ − 1 u a + C = 1 sin ⁡ θ tan ⁡ − 1 ( 1 − cos ⁡ θ sin ⁡ θ ) − 1 sin ⁡ θ tan ⁡ − 1 ( − cos ⁡ θ sin ⁡ θ ) = 1 sin ⁡ θ [ tan ⁡ − 1 ( 2 sin ⁡ 2 θ 2 2 sin ⁡ θ 2 cos ⁡ θ 2 ) + tan ⁡ − 1 ( cot ⁡ θ ) ] using cos ⁡ 2 u = 1 − 2 sin ⁡ 2 u and sin ⁡ 2 u = 2 sin ⁡ u cos ⁡ u = 1 sin ⁡ θ [ tan ⁡ − 1 ( tan ⁡ θ 2 ) + tan ⁡ − 1 ( tan ⁡ ( π 2 − θ ) ) ] = 1 sin ⁡ θ ( θ 2 + π 2 − θ ) = π − θ 2 sin ⁡ θ \begin{aligned}\int_0^1\frac1{x^2-2x\cos\theta+1}dx&=\int_0^1\frac1{(x-\cos\theta)^2+1-\cos^2\theta}dx\\ &=\int_0^1\frac1{\sin^2\theta+(x-\cos\theta)^2}dx\\ &=\left[\frac{1}{\sin\theta}\tan^{-1}\left(\frac{x-\cos\theta}{\sin\theta}\right)\right]_0^1&\small\text{using }\color{#3D99F6}\int\frac{1}{a^2+u^2}du=\frac{1}{a}\tan^{-1}\frac{u}{a}+C\\ &=\frac{1}{\sin\theta}\tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)-\frac{1}{\sin\theta}\tan^{-1}\left(\frac{-\cos\theta}{\sin\theta}\right)\\ &=\frac{1}{\sin\theta}\left[\tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)+\tan^{-1}\left(\cot\theta\right)\right]&\small\text{using }\color{#3D99F6}\cos2u=1-2\sin^2u\color{#333333}\text{ and }\color{#3D99F6}\sin2u=2\sin u\cos u\\ &=\frac{1}{\sin\theta}\left[\tan^{-1}\left(\tan\frac{\theta}{2}\right)+\tan^{-1}\left(\tan\left(\frac{\pi}{2}-\theta\right)\right)\right]\\ &=\frac{1}{\sin\theta}\left(\frac{\theta}{2}+\frac{\pi}{2}-\theta\right)\\ &=\boxed{\frac{\pi-\theta}{2\sin\theta}}\end{aligned}

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U forget a 2 (Pi- theta) /2sin(theta)

Ahpa Tsum - 3 years, 1 month ago

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Fixed, thanks for letting me know.

Joseph Newton - 3 years, 1 month ago

Let x 2 − 2 x cos ⁡ θ + 1 = u 2 x^2 - 2x \cos \theta + 1 = u^2 . Then x x , 1 1 and u u are the three sides of a triangle because they follow the cosine rule , where angle θ \theta is opposite side u u . Let the angle opposite x x be ϕ \phi . Then by sine rule , we have:

x sin ⁡ ϕ = u sin ⁡ θ = 1 sin ⁡ ( π − θ − ϕ ) = 1 sin ⁡ ( θ + ϕ ) \begin{aligned} \frac x{\sin \phi} & = \frac u{\sin \theta} = \frac 1{\sin (\pi - \theta - \phi)} = \frac 1{\sin (\theta + \phi)} \end{aligned}

x = sin ⁡ ϕ sin ⁡ ( θ + ϕ ) \begin{aligned} x & = \frac {\sin \phi}{\sin (\theta + \phi)} \end{aligned} , when x = 0 x = 0 , ϕ = 0 \phi = 0 and x = 1 x=1 , ϕ = π − θ 2 \phi = \dfrac {\pi - \theta}2 .

u = sin ⁡ θ sin ⁡ ( θ + ϕ ) \begin{aligned} u = \frac {\sin \theta}{\sin (\theta + \phi)}\end{aligned}

⟹ d x = cos ⁡ ϕ sin ⁡ ( θ + ϕ ) − sin ⁡ ϕ cos ⁡ ( θ + ϕ ) sin ⁡ 2 ( θ + ϕ ) d ϕ = sin ⁡ θ sin ⁡ 2 ( θ + ϕ ) d ϕ \begin{aligned} \implies dx = \frac {\cos \phi \sin (\theta + \phi) - \sin \phi \cos (\theta + \phi)}{\sin^2 (\theta + \phi)} d \phi = \frac {\sin \theta}{\sin^2 (\theta + \phi)} d \phi \end{aligned}

Therefore,

I = ∫ 0 1 1 x 2 − 2 x cos ⁡ θ + 1 d x = ∫ 0 1 1 u 2 d x = ∫ 0 π − θ 2 sin ⁡ 2 ( θ + ϕ ) sin ⁡ 2 θ ⋅ sin ⁡ θ sin ⁡ 2 ( θ + ϕ ) d ϕ = ∫ 0 π − θ 2 1 sin ⁡ θ d ϕ = π − θ 2 sin ⁡ θ \begin{aligned} I & = \int_0^1 \frac 1{x^2 - 2x \cos \theta + 1} dx \\ & = \int_0^1 \frac 1{u^2} dx \\ & = \int_0^{\frac {\pi - \theta}2} \frac {\sin^2 (\theta + \phi)}{\sin^2 \theta} \cdot \frac {\sin \theta}{\sin^2 (\theta + \phi)} d \phi \\ & = \int_0^{\frac {\pi - \theta}2} \frac 1{\sin \theta} d \phi \\ & = \boxed{\dfrac {\pi - \theta}{2\sin \theta}} \end{aligned}

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