121616949-math.179 - x Z p 1 x 2 dx = sec u tan u 2 ln |...

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8.3 Trigonometric Substitutions 165 Now for R sec 3 u du : sec 3 u = sec 3 u 2 + sec 3 u 2 = sec 3 u 2 + (tan 2 u + 1) sec u 2 = sec 3 u 2 + sec u tan 2 u 2 + sec u 2 = sec 3 u + sec u tan 2 u 2 + sec u 2 . We already know how to integrate sec u , so we just need the first quotient. This is “simply” a matter of recognizing the product rule in action: Z sec 3 u + sec u tan 2 u du = sec u tan u. So putting these together we get Z sec 3 u du = sec u tan u 2 + ln | sec u + tan u | 2 + C, and reverting to the original variable
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Unformatted text preview: x : Z p 1 + x 2 dx = sec u tan u 2 + ln | sec u + tan u | 2 + C = sec(arctan x ) tan(arctan x ) 2 + ln | sec(arctan x ) + tan(arctan x ) | 2 + C = x √ 1 + x 2 2 + ln | √ 1 + x 2 + x | 2 + C, using tan(arctan x ) = x and sec(arctan x ) = p 1 + tan 2 (arctan x ) = √ 1 + x 2 ....
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