Three Basic Substitutions

d dx ln 1+x √ 1−x2 = d dx ln(1+x)− 1 2 ln 1−x2 = 1 1+x − 1 2 −2x 1−x2 = 1−x 1−x2 + x 1−x2 = 1 1−x2 In practice we will usually need to construct a reference triangle as shown below. The substitution (8) sinθ =x = x 1 allows us to construct the (reference) triangle for θ. θ′ √ 1− x2 1 |x| We usually want the ... ................
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