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Question Number 105700 by john santu last updated on 31/Jul/20

\underset{- π}{\int^{π}} \frac{x^{2} d x}{1 + \sin (\sin x) + \sqrt{1 + \sin^{2} (\sin x)}}

Answered by bramlex last updated on 31/Jul/20

I = \underset{- π}{\int^{π}} \frac{x^{2} d x}{1 + \sqrt{1 + \sin^{2} (\sin x)} + \sin (\sin x)}

r e p l a c e x b y - x

I = \underset{π}{\int^{- π}} \frac{x^{2} (- d x)}{1 + \sqrt{1 + \sin^{2} (\sin x)} - \sin (\sin x)}

I = \underset{- π}{\int^{π}} \frac{x^{2} d x}{1 + \sqrt{1 + \sin^{2} (\sin x)} - \sin (\sin x)}

2 I = \underset{- π}{\int^{π}} \frac{2 x^{2} (1 + \sqrt{1 + \sin^{2} (\sin x)}) d x}{2 (1 + \sqrt{1 + \sin^{2} (\sin x)})}

2 I = \underset{0}{\int^{π}} \frac{4 x^{2} (1 + \sqrt{1 + \sin^{2} (\sin x)})}{2 (1 + \sqrt{1 + \sin^{2} (\sin x)})} d x

I = \underset{0}{\int^{π}} x^{2} d x = \frac{1}{3} π^{3} . ★

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