By Russell L. Herman

Creation 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 basic Harmonic OscillatorSecond Order Linear Differential EquationsLRC CircuitsDamped OscillationsForced SystemsCauchy-Euler EquationsNumerical options of ODEsNumericalRead more...

summary: advent and ReviewWhat Do i have to be aware of 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 suggestions of ODEsNumerical ApplicationsLinear SystemsProblemsLinear AlgebraFinite Dimensional Vector SpacesLinear TransformationsEigenvalue ProblemsMatrix formula of Planar SystemsApplicationsAppendix: Diagonali

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For example, consider cos3 x dx. This can be rewritten as cos3 x dx = Integration of odd powers of sine and cosine. cos2 x cos x dx. Let u = sin x. Then, du = cos x dx. Since cos2 x = 1 − sin2 x, we have cos3 x dx = cos2 x cos x dx = (1 − u2 ) du 1 = u − u3 + C 3 1 = sin x − sin3 x + C. 74) A quick check confirms the answer: d dx sin x − 1 sin3 x + C 3 = cos x − sin2 x cos x = cos x (1 − sin2 x ) = cos3 x. Integration of even powers of sine and cosine. 75) Even powers of sines and cosines are a little more complicated, but doable.

For example, the nth Root Test gives the convergence condition for an = |cn || x − a|n , √ ρ = lim n an = lim n |cn || x − a| < 1. n→∞ n→∞ Because | x − a| is independent of n, we can factor it out of the limit and divide the value of the limit to obtain −1 | x − a| < lim n→∞ n |cn | ≡ R. Thus, we have found the radius of convergence, R. Similarly, we can apply the Ratio Test: ρ = lim n→∞ a n +1 |c | = lim n+1 | x − a| < 1. n→∞ |cn | an introduction and review Again, we rewrite this result to determine the radius of convergence: | x − a| < | c n +1 | n→∞ |cn | lim −1 ≡ R.

12. Use differentiation under the integral sign to evaluate First, consider the integral I ( x, a) = Then, e ax dx = ∂I ( x, a) = ∂a e ax . a xe ax dx. xe x dx. introduction and review So, xe ax dx ∂I ( x, a) ∂a ∂ e ax dx ∂a ∂ e ax ∂a a x 1 − 2 e ax . 70) Evaluating this result at a = 1, we have xe x dx = ( x − 1)e x . The reader can verify this result by employing the previous methods or by just differentiating the result. I ( a) π ≡ 0 = = x2 cos x dx once more. 13. We will do the integral the integral cos ax dx sin ax π a 0 sin aπ .