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8.1 Integration by parts. Product Rule:. 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. OR. Let u be a portion of the integrand whose
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8.1 Integration by parts Product Rule:
Integration by parts Let dv be the most complicated part of the original integrand that fits a basic integration Rule (including dx). Then uwill be the remaining factors. OR 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).
Integration by parts u = x dv= exdx du = dx v = ex
Integration by parts u = lnx dv= x2dx du = 1/x dx v = x3 /3
Integration by parts v = x u = arcsin x dv= dx
Integration by parts u = x2 dv = sin x dx du = 2x dx v = -cos x u = 2xdv = cos x dx du = 2dx v = sin x
8.2 Trigonometric Integrals 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 anduse for all other factors of cos. 3. If neither power is odd, use power reducing formulas: Powers of Sine and Cosine
Tangents and secants Create an integral that is shown above.
8.3 Eliminating radicals by trig substitution. Pythagorean identities: Let u = a sin θ
Ex: Let x = a sin θ = 3 sin θ dx = 3 cos θdθ
Ex: Let u=2x, a=1 so 2x = tan θ dx = ½ sec2 θdθ
8.4 Partial Fractions If x = 2: 1=-B so B = -1 If x =3: 1=A
Partial Fractions-Repeated linear factors If x =0: 6= A If x = -1: -9 = -C, so C = 9 If x = 1: 31=6(4)+2B+9, B = - 1
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.
Repeated quadratic Factors A=8 For third degree: For second degree: B=0 13=2A+C For first degree: D+2B=0 For constant:
Repeated quadratic Factors A=8 B=0 13=2A+C D+2B=0 So, D=0 and C = -3
Improper Integrals with infinite limits Upper limit infinite Lower limit infinite Both limit infinite
Evaluation Use L’Hôpital’s rule We say the improper integral CONVERGES to The value of 1. (The area is finite.)
Improper Integrals-integrand becomes infinite interior point upper endpoint lower endpoint
Integrals with Infinite discontinuities. The integral converges to 2.
Area is finite Integral converges to 1
Area is infinite Integral diverges
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?
Direct comparison test Converges if Converges If f and g are continuous functions with f(x) g(x) For all x a. Then….. A function converges if its values are smaller than another function known to converge. Diverges if Diverges A function diverges if its values are larger than another function known to diverge.
Limit Comparison test for convergence If f and g are positive and continuous on [a, ) And if and Then the integrals both converge or both diverge: If diverges and If then also diverges.