121.pdf - Let x = 9 tan \u03b8 where \u2212 \u03c02 < \u03b8 < \u03c02 Then dx = 9 sec2 \u03b8 d\u03b8 and p \u221a \u221a \u221a x2 81 = 81 tan2 \u03b8 81 = 81(tan2 \u03b8 1 = 81 sec2 \u03b8 = 9 |sec

121.pdf - Let x = 9 tan u03b8 where u2212 u03c02 <...

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Unformatted text preview: Let x = 9 tan θ, where − π2 < θ < π2 . Then dx = 9 sec2 θ dθ and p √ √ √ x2 + 81 = 81 tan2 θ + 81 = 81(tan2 θ + 1) = 81 sec2 θ = 9 |sec θ| = 9 sec θ for the relevant values of θ. 3 R 729 tan3 θ R R x √ dx = 9 sec2 θdθ = 729 tan3 θ sec θdθ 9 sec θ x2 + 81 R = 729 tan2 θ tan θ sec θdθ R = 729 (sec2 θ − 1) tan θ sec θdθ R = 729 (u2 − 1)du [u = sec θ, du = sec θ tan θdθ]   = 729 13 u3 − u + C = 729 31 sec3 θ − sec θ + C " # √ 1 (x2 + 81)3/2 x2 + 81 = 729 − +C 3 729 9 √ = 13 (x2 + 81)3/2 − 81 x2 + 81 + C b=9 ...
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