# 8.5 - 8.5 1 8.5 Trigonometric Substitution Three Basic...

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Unformatted text preview: 8.5 1 8.5 Trigonometric Substitution Three Basic Substitutions Recall the derivative formulas for the inverse trigonometric functions of sine, secant, tangent. d dx ( sin − 1 x ) = 1 √ 1 − x 2 , | x | < 1 (1) d dx ( tan − 1 x ) = 1 1 + x 2 (2) d dx ( sec − 1 x ) = 1 | x | √ x 2 − 1 , | x | > 1 (3) Now focus on the quadratic term in each derivative. 8.5 2 For example, in (1), we focus on the 1 − x 2 under the radical. 1 √ 1 − x 2 (4) Is there a substitution that can help us with this? Let’s try x = sin θ . Thus sin θ = x 1 (5) Now we can construct the (reference) triangle for θ . Thus θ ′ √ 1 − x 2 1 x It follows that 1 √ 1 − x 2 = | sec θ | In practice, we usually want the substitution in (5) to be invertible. That is, x = sin θ iff θ = sin − 1 x, − π 2 ≤ θ ≤ π 2 (6) 8.5 3 Example 1. Trig Substitution using x = sin θ Evaluate (7) integraldisplay 1 1 − x 2 dx If (8) − 1 < x < 1 we can let x = sin θ, − π/ 2 ≤ θ ≤ π/ 2 . Then dx = cos θ dθ So that integraldisplay dx 1 − x 2 = integraldisplay cos θ dθ cos 2 θ = integraldisplay sec θ dθ = ln | sec θ + tan θ | + C = ln vextendsingle vextendsingle vextendsingle vextendsingle...
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8.5 - 8.5 1 8.5 Trigonometric Substitution Three Basic...

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