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Unformatted text preview: = tan 2 t sec t2 Z u 2 du = tan 2 t sec t2 u 3 3 + C = tan 2 t sec t2 sec 3 t 3 + C. Therefore Z x 3 dx √ x 2 + 1 dx = tan 2 t sec t2 sec 3 t 3 + C. 2. Back substitute to the original variable. Using the original trig substitution x = tan t , we have t = arctan x so that sec t = sec (arctan x ) = √ 1 + x 2 (drawing a right triangle). Thus, Z x 3 dx √ x 2 + 1 dx = x 2 √ 1 + x 22(1 + x 2 ) 3 / 2 3 + C = √ 1 + x 2 3 ± x 22 ² + C. 2...
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 Fall '10
 KIHYUNHYUN
 Calculus, sec t dt

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