Unformatted text preview: = 3 cos θdθ. Z 1 (54 xx 2 ) 3 / 2 dx = Z 1 (99 sin 2 θ ) 3 / 2 (3 cos θ ) dθ = Z 3 cos θ (3 cos θ ) 3 dθ = 1 9 Z 1 cos 2 θ dθ = 1 9 Z sec 2 θdθ = 1 9 tan θ + C. Finally, since sin θ = x +2 3 , tan θ = x +2 √ 54 xx 2 , and Z 1 (54 xx 2 ) 3 / 2 dx = x + 2 9 √ 54 xx 2 + C. 1...
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 Fall '06
 McAdam
 Calculus, Following, tan θ, sec x tan, tan xdx

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