Section8_3_review

Section8_3_review - Section 8.3 Trigonometric Substitution...

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Section 8.3: Trigonometric Substitution Please refer to handout for review material. Example. Evaluate x 2 9 - x 2 dx . Solution. Since the radical contains a 2 - x 2 , let x = 3sin θ , - π 2 ≤θ≤ 2 dx = 3cos d x 2 = 9sin 2 9 - x 2 = 9 - 9sin 2 = 91 - sin 2 ( 29 = 9cos 2 = 3cos Therefore, x 2 9 - x 2 dx = 9sin 2 3cos 3cos d = 9 sin 2 d = 9 1 2 1 - cos2 ( 29 d = 9 2 d θ- 9 2 cos2 d = 9 2 9 2 1 2 sin2 + c And s in2 θ= 2s in cos = 9 2 9 2 1 2 2sin cos + c = 9 2 9 2 sin cos + c Now , sin and cos need to be expressed in terms of the original variable x . Since x = 3sin , sin x 3 = opposite hypotenuse and = sin - 1 x 3 Drawing a triangle and labeling the sides, We can determine cos = adjacent hypotenuse , or cos 9 - x 2 3 Finally,
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x 2 9 - x 2 dx ò = 9 2 sin - 1 x 3 - 9 2 × x 3 × 9 - x 2 3 + c = 9 2 sin - 1 x 3 - x 2 9 - x 2 + c Example. Evaluate 1 25 x 2 - 4 dx ò . Solution. Adjust the radical before making the substitution, 1 25 x 2 - 4 dx ò = 1 25 x 2 - 4 25 ae è ç ö ø ÷ ò dx = 1 5 x 2 - 4 25 dx = 1 5 1 x 2 - 4 25 dx ò ò Since the radical contains x 2 - a 2 , let x = 2 5 sec
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This note was uploaded on 01/12/2010 for the course MATH 44245 taught by Professor Famiglietti during the Fall '07 term at UC Irvine.

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Section8_3_review - Section 8.3 Trigonometric Substitution...

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