4 Closed-Form Solutions for the Equations of Motion of the Heavy Symmetrical Top with One Point Fixed 33 The differential of [x2 =h(1−u)], results in dx = −h 2x du. (4.23) Replacing (4.22) and (4.23) in the right-hand side of (4.21) results in −1 −h 2x du 1 x √uh −(h−1) . (4.24) Further manipulation produces 1 2 1 (1−u) du √h u − h−1 h , (4.25) and using the definition (4.18), the right-hand side of (4.21) is converted to 1 2√h 1 (1−u) du √u−u0 . (4.26) Replacing the result of (4.20) and (4.26) into the full expression of (4.21): d sech−1 x = −1 x √1 −x2 dx = 1 2 √h 1 (1−u) du √u−u0 = 1 2 √h β t. (4.27) Replacing (4.22) in (4.27) is obtained by the following equation: sech−1 h (1−u) −sech−1 h (1−u0) = 1 2 β h t. (4.28) Because sech−1 √h (1 −u0) =0, (4.28) is further reduced to sech−1 h (1−u) = 1 2 β h t. (4.29) Solving for the inner term in sech−1, the following is obtained: h (1−u) =sech 2 1 2 β h t . (4.30) The following trigonometric identities will help to reduce (4.30):
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