0-pi-x-sin-x-1-cos-2-x-dx-

Question Number 113738 by bemath last updated on 15/Sep/20

\underset{0}{\int^{π}} \frac{x \sin x}{1 + \cos^{2} x} d x ?

Answered by bobhans last updated on 15/Sep/20

I = \underset{0}{\int^{π}} \frac{x \sin x}{1 + \cos^{2} x} dx

replace x by π - x \to I = \underset{π}{\int^{0}} \frac{(π - x) \sin (π - x)}{1 + \cos^{2} (π - x)} (- dx)

I = \underset{0}{\int^{π}} \frac{(π - x) \sin x}{1 + \cos^{2} x} dx = \underset{0}{\int^{π}} \frac{π \sin x}{1 + \cos^{2} x} dx - \underset{0}{\int^{π}} \frac{xsin x}{1 + \cos^{2} x} dx

2 I = \underset{0}{\int^{π}} \frac{π \sin x}{1 + \cos^{2} x} dx

consider \int \frac{π \sin x}{1 + \cos x} dx = - π \int \frac{d (\cos x)}{1 + \cos^{2} x}

= - π \int \frac{du}{1 + u^{2}}

= - π \tan^{- 1} (\cos x) + c

now we have 2 I = - π {[\tan^{- 1} (\cos x)]}_{0}^{π}

I = - \frac{π}{2} [- \frac{π}{4} - \frac{π}{4}] = \frac{π^{2}}{4} .

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