Using half-angle substitution, if we let z = tan z, the f dx 1+sin x+cos X...

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Using half-angle substitution, if we let z = tan z, the f dx 1+sin x+cos X in terms of z, is equivalent to A. Sdz 1+22 B. S dz c.S 1+z D. Sd 2+z2 Using half-angle substitution, if we let z = tan z, the f dx 1+sin x+cos X in terms of z, is equivalent to A. Sdz 1+22 B. S dz c.S 1+z D. Sd 2+z2 Using half-angle substitution, if we let z = tan z, the f dx 1+sin x+cos X in terms of z, is equivalent to A. Sdz 1+22 B. S dz c.S 1+z D. Sd 2+z2 Using half-angle substitution, if we let z = tan z, the f dx 1+sin x+cos X in terms of z, is equivalent to A. Sdz 1+22 B. S dz c.S 1+z D. Sd 2+z2