149x Filetype PDF File size 0.05 MB Source: lagrange.math.siu.edu
Math 305 Methods of Integration The following are a list of integration formulas that you should know. Even when you go through the different methods, you reduce the integrals to one of these: Z n+1 Z n x 2 x dx = n+1+cn6=−1 sec xdx = tanx+c Z dx Z 2 x =ln|x|+c csc xdx = −cotx+c Z x x Z e dx = e +c secxtanxdx = secx+c Z sinxdx =−cosx+c Z cscxcotxdx = −cscx+c Z cosxdx = sinx+c The following are nice to know (make your life easier) integrals: Z tanxdx =−ln|cosx|+c Z cscxdx = −ln|cscx+cotx|+c Z cotxdx =ln|sinx|+c Z secxdx =ln|secx+tanx|+c If a is constant, a 6=0: Z ax 1 ax Z 1 Z 1 e dx = ae +c cosaxdx = a sinax+c sinaxdx = −a cosax+c HintI:Ifyouthinkyouhaveanantiderivative,justdifferentiateandseeifyouendupwiththeintegrand! Hint II: The line of attack you should take when trying to integrate should be to try these methods in this order. You will always need to have paper and pencil, the correct method doesn’t usually bounce off the page and hit you in the face! Math 305 MI-Page 2 Method 1. Straight forward integration. The integral can be reduced to one of the first set of integrals. Z Z 4 2 2 3 2 x 3 x Example A. (x +1)(x−3)dx = (x −3x +x−3)dx= 4 −x + 2 −3x+c Z 2 Z Z Example B. 1−sin xdx = (1 −sinx)(1+sinx)dx = (1 −sinx)dx = x+cosx+c 1+sinx 1+sinx Problems: Z x −x 2 1. e (1−e sec x)dx Z x3−2x+4 2. x dx 3. Z cosxtanxdx Method 2. Substitution. A simple substitution reduces the integral to one of the first set of integrals. In fact, the second set came about by using substitution. Example C. Z tanxdx = Z sinxdx. Let w = cosx then dw = −sinxdx so that the integral becomes Z cosx −dw =−ln|w|+c=−ln|cosx|+c. w Z 2 Z dw 1 x 2 w w Example D. xe dx. Let w = x then dw =2xdx so that the integral becomes e 2 = 2e +c= 1 2 ex +c. 2 Hint III: Things to look for: if the integrand involves f(x) 1 n e ; trig (f(x)); f(x); (f(x)) Let w = f(x): This is not an exclusive list! Hint IV: When substitution is complete, make sure no x’s appear in the integrand. Math 305 MI-Page 3 Problems: 4. Z cosx dx. Let w=1+sinx: 1+sinx Z 2 3 2 3 2 5. (3x +x)cos(2x +x +4)dx. Let w =2x +x +4: 6. Z (lnx)3dx. Let w =lnx. x Z 2 tanx 7. sec xe dx Z √ 8. x x+2dx 9. Z 2 x+1 dx x +2x−5 Method 3. Integration by parts. This method undoes the product rule Z udv = uv−Z vdu Z 2 Example E. xlnxdx: Let u =lnx; dv = xdx then du = dx and v = x . x 2 Z 2 Z 2 2 2 xlnxdx =(lnx)x − x dx = x lnx− x +c 2 2 x 2 4 Example F. Z xsinxdx: Let u = x; dv = sinxdx: Then dv = dx and v = −cosx. Z xsinxdx = x(−cosx)−Z −cosxdx =−xcosx+sinx+c Math 305 MI-Page 4 Hint V: When choosing u and dv make sure dv is something that can be integrated. Also, the whole integrand should be taken up with u and dv. Hint VI: Method should be used when integrand involves ax ax (2n+1) (poly)trig e (sinbx) e cosbx sec x (2n+1) (poly) lnx poly(inverse trig fact) csc x This is not an exclusive list! Problems: 10. Z lnxdx. Let u =lnx; dv = dx: Z 2 2 11. (x +2x−1)cos3xdx. Let u=x +2x−1;dv=cos3xdx: (Need to use integration by parts twice.) Z 2x 2x 12. (x +3)e dx. Let u=x+3;dv=e dx: Z 2 13. x lnxdx Z −1 14. tan xdx Z x 15. e sinxdx Method 4. Trigonometric integrals. Integrands only involve trigonometric functions (not inverse trig functions!). Remember that certain trig functions “go together.” sinθ and cosθ tanθ and secθ cotθ and cotθ If you have mixed trig functions convert everything to sinθ and cosθ. Another helpful identity is 2 2 2 2 2 2 sin θ+cos θ =1: From here you can derive tan θ+1 = sec θ and 1+cot θ = csc θ: You should also have the half angle formulas in your repertoire. 2 1 sin θ =2(1−cos2θ) 2 1 cos θ =2(1+cos2θ)
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