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$The value of I = \int_{0}^{2 π} (\frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x}) d x is equal to$

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2π2

π2

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detailed solution

Correct option is B

2I=∫02πxsin8xsin8x+cos8x+(2π-x)sin8xsin8x+cos8xdx=∫0π4πsin8xsin8x+cos8xdx=∫0π/28πsin8xsin8x+cos8xdx=4π∫0π/2sin8xsin8x+cos8x+cos8xsin8x+cos8xdx=4π·π2=2π2 Therefore, I=π2

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The value of I=∫02πxsin8xsin8x+cos8xdx is equal to

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$The value of I = \int_{0}^{2 π} (\frac{x \sin^{8} x}{\sin^{8} x + \cos^{8} x}) d x is equal to$