A Nice Integral (3)

Calculus Level 3

Evaluate the following integral ∫ 0 π x sin ⁡ x 1 + cos ⁡ 2 x d x \large \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x} \, dx

π 4 \frac{\pi}{4} π 2 2 \frac{\pi^2}{2} π 3 \frac{\pi}{3} π 2 4 \frac{\pi^2}{4}

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Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
Aug 11, 2016

I = ∫ 0 π x sin ⁡ x 1 + cos ⁡ 2 x d x Using the identity ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x = 1 2 ∫ 0 π x sin ⁡ x 1 + cos ⁡ 2 x + ( π − x ) sin ⁡ ( π − x ) 1 + cos ⁡ 2 ( π − x ) d x = 1 2 ∫ 0 π x sin ⁡ x 1 + cos ⁡ 2 x + ( π − x ) sin ⁡ x 1 + cos ⁡ 2 x d x = 1 2 ∫ 0 π π sin ⁡ x 1 + cos ⁡ 2 x d x Let u = cos ⁡ x ⟹ d u = − sin ⁡ x d x = 1 2 ∫ − 1 1 π 1 + u 2 d u = π tan ⁡ − 1 u 2 ∣ − 1 1 = π 2 4 \begin{aligned} I & = \int_0^\pi \frac {x\sin x}{1+\cos^2 x}dx \quad \quad \small \color{#3D99F6}{\text{Using the identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^\pi \frac {x\sin x}{1+\cos^2 x} + \frac {(\pi-x)\sin (\pi-x)}{1+\cos^2 (\pi-x)} \ dx \\ & = \frac 12 \int_0^\pi \frac {x\sin x}{1+\cos^2 x} + \frac {(\pi-x)\sin x}{1+\cos^2 x} \ dx \\ & = \frac 12 \int_0^\pi \frac {\pi \sin x}{1+\cos^2 x} dx \quad \quad \small \color{#3D99F6}{\text{Let }u = \cos x \implies du = - \sin x \ dx} \\ & = \frac 12 \int_{-1}^1 \frac \pi{1+u^2} du \\ & = \frac {\pi \tan^{-1} u}2\bigg|_{-1}^1 \\ & = \boxed{\dfrac {\pi^2} 4} \end{aligned}

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Thank you for the nice solution.

Hana Wehbi - 4 years, 10 months ago

It should be π × tan ⁡ − 1 u 2 ∣ − 1 1 = π 2 4 \pi \times \dfrac{\tan^{-1} u}{2} \bigg|_{-1}^{1} = \dfrac{\pi^2}{4} .

Tapas Mazumdar - 4 years, 3 months ago

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Thanks. I got it changed.

Chew-Seong Cheong - 4 years, 3 months ago

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