$ \int { { {\cos 2x} \over {\sin x - \cos x}}dx} = $( ) A: $\sin x + \cos x + C $ B: $ - \sin x + \cos x + C $ C: $ - \sin x- \cos x + C $ D: $ \sin x - \cos x + C $

2022-06-06

$ \int { { {\cos 2x} \over {\sin x - \cos x}}dx} = $( ) A: $\sin x + \cos x + C $ B: $ - \sin x + \cos x + C $ C: $ - \sin x- \cos x + C $ D: $ \sin x - \cos x + C $

$ \int { { {\cos 2x} \over {\sin x - \cos x}}dx} = $( )
A: $\sin x + \cos x + C $
B: $ - \sin x + \cos x + C $
C: $ - \sin x- \cos x + C $
D: $ \sin x - \cos x + C $

答案：

B

举一反三

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\( \int { { {\cos 2x} \over {\sin x - \cos x}}dx} = \)( ) A: \(\sin x + \cos x + C \) B: \( - \sin x + \cos x + C \) C: \( - \sin x- \cos x + C \) D: \( \sin x - \cos x + C \)

举一反三

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