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∫sen xdx√sin (2x+A)+sinA=∫sec xdx√2 sin x cosx+(2cos2x−1)sin A+sinA=∫sec x dx√2 cos x sin (x+A)\

Answer» ∫sen xdx√sin (2x+A)+sinA=∫sec xdx√2 sin x cosx+(2cos2x−1)sin A+sinA=∫sec x dx√2 cos x sin (x+A)\


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