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By Dr. Safa Ahmed El-Askary Faculty of Allied Medical of Sciences Lecture (7&8) Integration by Parts 1
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Integration by parts Product Rule:
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Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule (including dx). Then u will be the remaining factors. Let u be a portion of the integrand whose derivative is a function simpler than u. Then dv will be the remaining factors (including dx). OR
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Integration by parts u = x dv= e x dx du = dx v = e x
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Integration by parts u = lnx dv= x 2 dx du = 1/x dx v = x 3 /3
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Integration by parts u = arcsin x dv= dx v = x
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Integration by parts u = x 2 dv = sin x dx du = 2x dx v = -cos x u = 2x dv = cos x dx du = 2dx v = sin x
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8.2 Trigonometric Integrals Powers of Sine and Cosine 1. If n is odd, leave one sin u factor and use for all other factors of sin. 2. If m is odd, leave one cos u factor and use for all other factors of cos. 3. If neither power is odd, use power reducing formulas:
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Powers of sin and cos
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Tangents and secants Create an integral that is shown above.
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8.3 Eliminating radicals by trig substitution. Pythagorean identities: Let u = a sin θ
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Trig Substitutions
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Let x = a sin θ = 3 sin θ dx = 3 cos θ dθ Ex:
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Let u=2x, a=1 so 2x = tan θ dx = ½ sec 2 θ dθ Ex:
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8.4 Partial Fractions If x = 2: 1=-B so B = -1 If x =3: 1=A
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Partial Fractions-Repeated linear factors If x =0: 6= A If x = 1: 31=6(4)+2B+9, B = - 1 If x = -1: -9 = -C, so C = 9
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Quadratic Factors If x = 0 then A = 2 If x = 1 then B = -2 If x = -1 2 = -C +D If x = 2 8 = 2C+D Solving the system of equations you find C = 2 and D = 4.
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Repeated quadratic Factors A=8 13=2A+C For third degree:For second degree: B=0 For first degree: For constant: D+2B=0
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Repeated quadratic Factors A=8 13=2A+C B=0 D+2B=0 So, D=0 and C = -3
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8.8 Areas under curves with infinite domain or range
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Improper Integrals with infinite limits Upper limit infinite Lower limit infinite Both limit infinite
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Infinite limits
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We say the improper integral CONVERGES to The value of 1. (The area is finite.) Evaluation Use L’Hôpital’s rule
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When both limits are infinite
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Improper Integrals-integrand becomes infinite upper endpoint lower endpoint interior point
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Integrals with Infinite discontinuities. The integral converges to 2.
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Calculation with infinite discontinuity
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Area is finite Integral converges to 1
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Area is infinite Integral diverges
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Integrals of the form
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Convergence or divergence Integrals of the form Converge if p > 1 and diverge if p = 1 or p < 1. Which of the following converge and which diverge?
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Direct comparison test If f and g are continuous functions with f(x) g(x) For all x a. Then….. Converges ifConverges Diverges ifDiverges A function converges if its values are smaller than another function known to converge. A function diverges if its values are larger than another function known to diverge.
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Limit Comparison test for convergence If f and g are positive and continuous on [a, ) And if Then the integrals and If both converge or both diverge: diverges and thenalso diverges.
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