結果 : is sec^2x = 1/cos^2x
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If ∫√sec 2x-1dx= loge|cos 2x + β +√cos 2x (1+cos 1/β x| + constant, then β- α is equal to

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(1/sec^2x-cos^2x + 1/cosec^2x-sin^2x)sin^2xcos^2x =| 1-sin^2xcos^2x/2+sin^2xcos^2x

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