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Help please! How to evaluate the integral [sqrt (cot x)] dx?

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Help please! How to evaluate the integral [sqrt (cot x)] dx?

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  1. if x is in (0, π/4] you could try u² = cot x, dx = -2u du / [u^4+1]....csc ² x = cot² x + 1...need to write a series for 1 / [ 1 + u^4] and then integrate term by term...2nd or 3rd term of calc?

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