How to find the indefinite integral of DX / (sin ^ 6x + cos ^ 6x) The problem has not been solved for a long time,

(secx) ^ 2DX = D (TaNx), (secx) ^ 4 = [1 + (TaNx) ^ 2] ^ 2 ∫ DX / (sin ^ 6x + cos ^ 6x) = ∫ [1 + (TaNx) ^ 2] ^ 2D (TaNx) / [1 + (TaNx) ^ 6] let t = TaNx = ∫ [1 + 2T ^ 2 + T ^ 4] DT / [1 + T ^ 6] = ∫ [1 - t

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