By Russell L. Herman

Advent and ReviewWhat Do i must be aware of From Calculus?What i want 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 have to 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 easy Harmonic OscillatorSecond Order Linear Differential EquationsLRC CircuitsDamped OscillationsForced SystemsCauchy-Euler EquationsNumerical options of ODEsNumerical ApplicationsLinear SystemsProblemsLinear AlgebraFinite Dimensional Vector SpacesLinear TransformationsEigenvalue ProblemsMatrix formula of Planar SystemsApplicationsAppendix: Diagonali

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Sin π 12 = = = Double angle formulae. π π − 3 4 π π π π sin cos − sin cos 3√ 4√ 4 3 √ 3 2 21 − 2 2 2 2 √ 2 √ 3−1 . 20) cos(2A) = cos2 A − sin2 A. 21) as cos(2A) = 2 cos2 A − 1, 2 = 1 − 2 sin A. Half angle formulae. 23) These, in turn, lead to the half angle formulae. Solving for cos2 A and sin2 A, we find that 1 − cos 2A , 2 1 + cos 2A cos2 A = . 2. Evaluate cos 12 . In the previous example, we used the sum/difference identities to evaluate a similar expression. We could have also used a half angle identity.

We let u = x2 + 1. As x goes from 0 to 2, u takes values from 1 to 5. So, this substitution gives 2 0 √ x x2 + 1 dx = 5 1 2 1 √ √ du √ = u|51 = 5 − 1. u When you become proficient at integration, you can bypass some of these steps. In the next example we try to demonstrate the thought process involved in using substitution without explicitly using the substitution variable. 6. Evaluate dx As with the previous example, one sees that the derivative of 9 + 4x2 is proportional to x, which is in the numerator of the integrand.

Since |r|<1, the geometric series converges. So, the sum of the series is given by S= 4 9 1 3 1− 2 = . 27. ∑∞ n =1 ( 2n − 5n ) Finally, in this case we do not have a geometric series, but we do have the difference of two geometric series. Of course, we need to be careful whenever rearranging infinite series. In this case it is allowed 1 . Thus, we have ∞ ∑ n =1 ∞ 3 2 − n 2n 5 ∞ 3 2 − . ∑ n n 2 5 n =1 n =1 ∑ = Now we can add both geometric series to obtain ∞ ∑ n =1 3 2 − n n 2 5 = 3 2 1− − 1 2 2 5 1− 1 5 = 3− 1 5 = .