We use integration by parts let u tan and dv sec tan

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Unformatted text preview: l problem is equal to tan2 θ sec θdθ. We use integration by parts. Let u = tan θ and dv = sec θ tan θdθ . It follows that du = sec2 θdθ and v = sec θ . So tan2 θ sec θdθ = sec θ tan θ − sec3 θdθ = sec θ tan θ − = sec θ tan θ − Add sec θdθ − (tan2 θ + 1) sec θdθ tan2 θ sec θdθ. tan2 θ sec θdθ to both sides to see that 2 tan2 θ sec θdθ = sec θ tan θ − sec θdθ. So x2 + 2xdx = tan2 θ sec θdθ = (1/2) sec θ tan θ − sec θdθ = (1/2) [sec θ tan θ − ln | sec θ + tan θ |] + C = (1/2) (x + 1) x2 + 2x − ln |(x + 1) + x2 + 2 x| + C . Check. The derivative of (1/2) (x + 1) x2 + 2x − ln[(x + 1) + is (x + 1)(2x + 2) √ (1/2) + 2 x2 + 2 x (x + 1)2 + = (1/2) √ x2 x2 + 2x − + 2x − x2 + 2 x 1+ (x + 1) 1+ 2 √ x+2 2 x2 +2x √ + x2 + √ x+1 x2 +2x √ 2x 2 (x + 1)2 + = (1/2) √ x2 + 2 x x2 + 2 x − (x + 1)2 + = (1/2) √ x2 + 2 x √ x2 + 2 x + x + 1 √ √ [(x + 1) + x2 + 2x] x2 + 2x x2 + 2 x − √ 1 x2 + 2 x 1 =√ (x + 1)2 + x2 + 2x − 1 2 + 2x 2x 1 =√ 2 x2 + 4 x 2 + 2x 2x =√ 1 x2 + 2 x = 2 + 2x x x2 + 2x....
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This document was uploaded on 03/23/2014 for the course MATH 142 at South Carolina.

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