# 8.5.15-eg - = tan 2 t sec t-2 Z u 2 du = tan 2 t sec t-2 u...

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Example Evaluate Z x 3 dx x 2 + 1 dx . 1. Use a trigonometric substitution We make the trig substitution x = tan t, dx = sec 2 t dt. Since the integrand is well-defined for all x , it follows that for all - π 2 t π 2 , we rewrite the integral as Z x 3 dx x 2 + 1 dx = Z tan 3 t sec 2 t dt sec 2 t = Z tan 3 t sec t dt where we have used that sec 2 t = | sec t | = sec t 0 for all - π 2 t π 2 . We view the integral as Z tan 3 t sec t dt = Z tan 2 t tan t sec t dt, and integrate by parts letting u = tan 2 t dv = tan t sec t dt du = 2 tan t sec 2 t dt v = sec t so that Z tan 3 t sec t dt = tan 2 t sec t - 2 Z tan t sec 3 t dt = tan 2 t sec t - 2 Z tan t sec t sec 2 t dt Making the substitution u = sec t du = tan t sec t , Z tan 3 t sec t dt = tan 2 t sec t - 2 Z tan t sec t

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Unformatted text preview: = tan 2 t sec t-2 Z u 2 du = tan 2 t sec t-2 u 3 3 + C = tan 2 t sec t-2 sec 3 t 3 + C. Therefore Z x 3 dx √ x 2 + 1 dx = tan 2 t sec t-2 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 2-2(1 + x 2 ) 3 / 2 3 + C = √ 1 + x 2 3 ± x 2-2 ² + C. 2...
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