Chapter8-T.pdf - 8 TECHNIQUES OF INTEGRATION OVERVIEW The...

This preview shows page 1 - 4 out of 61 pages.

8 TECHNIQUES OF INTEGRATION OVERVIEW The Fundamental Theorem tells us how to evaluate a definite integral once we have an antiderivative for the integrand function. Table 8.1 summarizes the forms of anti- derivatives for many of the functions we have studied so far, and the substitution method helps us use the table to evaluate more complicated functions involving these basic ones. In this chapter we study a number of other important techniques for finding antiderivatives (or indefinite integrals) for many combinations of functions whose antiderivatives cannot be found using the methods presented before. TABLE 8.1 Basic integration formulas 6. I sinxdx = -cosx + C 7. I cosxdx = sinx + C 8. I sec 2 xdx = tanx + C 9. I csc 2 xdx = -cotx + C 10. I secxtanxdx = secx + C 11. I cscxcotxdx = -cscx + C (x> a > 0) (a> 0) 21. 22. 12. I tanxdx = In Isecxl + C 13. I cotxdx = In Isinxl + c 14. I secxdx = In Isecx + tanxl + C 15. I cscxdx = -In Icscx + cotxl + C 16. I sinh x dx = coshx + C 17. I coshxdx = sinhx + C 18. I v'ar:- x 2 = sin- 1 (~) + C 19. I a2 ~ x 2 = ~tan-l (~) + C 20. I V dx = ~sec-l I ~ I + C X x 2 - a 2 (n =i' -1) (a>O,a=i' 1) (any number k) 3. I~ = In Ixl + C 4. I eX dx = eX + C 5. lax dx = L + C Ina 1. I kdx = kx + C 435
Image of page 1

Subscribe to view the full document.

436 Chapter 8: Techniques of Integration 8.1 Integration by Parts Integration by parts is a technique for simplifying integrals of the fann J j(x)g(x) dx. It is useful when j can be differentiated repeatedly and g can be integrated repeatedly without difficulty. The integrals JxCOSXdx and are such integrals because j(x) = x or j(x) = x 2 can be differentiated repeatedly to be- come zero, and g(x) = cos x or g(x) = eX can be integrated repeatedly without difficulty. Integration by parts also applies to integrals like J Inxdx and J eX cosxdx. In the rlISt case,j(x) = In x is easy to differentiate and g(x) = I easily integrates to x. In the second case, each part of the integrand appears again after repeated differentiation or inte- gration. Product Rule in Integral Fonn If j and g are differentiable functions of x, the Product Rule says that d dx [f(x)g(x)] = j'(x)g(x) + j(x)g'(x). In terms of indermite integrals, this equation becomes J :x [f(x)g(x)] dx = J [f'(x)g(x) + j(x)g'(x)] dx or J :x [f(x)g(x)] dx = J j'(x)g(x) dx + J j(x)g'(x) dx. Rearranging the terms of this last equation, we get J j(x)g'(x) dx = J :x [f(x)g(x)] dx - J j'(x)g(x) dx, leading to the integration by parts formula J j(x)g'(x) dx = j(x)g(x) - J j'(x)g(x) dx (l) Sometimes it is easier to remember the formula if we write it in differential form. Let u = j(x) and v = g(x). Then du = j'(x) dx and dv = g'(x) dx. Using the Substitution Rule, the integration by parts fannula becomes
Image of page 2
8.1 Integration by Parts 437 Integration by Parts Formula j udv = uv - j vdu (2) This formula expresses one integral, J u dv, in terms of a second integral, J v du. With a proper choice of u and v, the second integral may be easier to evaluate than the first. In using the formula, various choices may be available for u and dv.
Image of page 3

Subscribe to view the full document.

Image of page 4
  • Fall '11
  • Staff

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern