By Russell L. Herman

Creation and ReviewWhat Do i have to comprehend From Calculus?What i would like From My Intro Physics Class?Technology and TablesAppendix: Dimensional AnalysisProblemsFree Fall and Harmonic OscillatorsFree FallFirst Order Differential EquationsThe uncomplicated Harmonic OscillatorSecond Order Linear Differential EquationsLRC CircuitsDamped OscillationsForced SystemsCauchy-Euler EquationsNumerical recommendations of ODEsNumericalRead more...

summary: creation and ReviewWhat Do i must recognize From Calculus?What i would like From My Intro Physics Class?Technology and TablesAppendix: Dimensional AnalysisProblemsFree Fall and Harmonic OscillatorsFree FallFirst Order Differential EquationsThe basic Harmonic OscillatorSecond Order Linear Differential EquationsLRC CircuitsDamped OscillationsForced SystemsCauchy-Euler EquationsNumerical recommendations of ODEsNumerical ApplicationsLinear SystemsProblemsLinear AlgebraFinite Dimensional Vector SpacesLinear TransformationsEigenvalue ProblemsMatrix formula of Planar SystemsApplicationsAppendix: Diagonali

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Here we will use the substitution x = cosh u, dx = sinh u du, x2 − 1 = cosh2 u − 1 = sinh u. Then, √ dx x2 − 1 = sinh u du sinh u = du = u + C = cosh−1 x + C 1 = ln( x + x2 − 1) + C, x ≥ 1. 84) 2 This is the same result as we had obtained previously, but this derivation was a little cleaner. Also, we can extend this result to values x ≤ −1 by letting x = − cosh u. This gives dx 1 √ = ln( x + x2 − 1) + C, x ≤ −1. 2 2 x −1 Combining these results, we have shown √ dx x2 −1 = 1 ln(| x | + 2 x2 − 1) + C, x2 ≥ 1.

Periodic functions. These boxed equations are the most common trigonometric identities. They appear often and should just roll off your tongue. We will also need to understand the behaviors of trigonometric functions. In particular, we know that the sine and cosine functions are periodic. They are not the only periodic functions, as we shall see. ] However, they are the most common periodic functions. A periodic function f ( x ) satisfies the relation f ( x + p ) = f ( x ), for all x introduction and review for some constant p.

Also, we can use the identity cosh2 u − sinh2 u = 1 to rewrite x2 + 4 = 4 sinh2 u + 4 = 2 cosh u. The integral can be now be evaluated using these substitutions and some hyperbolic function identities, 2 x2 + 4 dx 0 = 4 = 2 sinh−1 1 0 sinh−1 1 0 cosh2 u du (1 + cosh 2u) du 1 2 u + sinh 2u 2 = sinh−1 1 0 = 2 [u + sinh u cosh u]sinh 0 √ −1 = 2 sinh 1 + 2 . 17 we used a trigonometric substitution and found 2 0 √ √ x2 + 4 = 2( 2 + ln( 2 + 1)). This is the same result since sinh−1 1 = ln(1 + √ 2). 21.