121616949-math.182 - x 2 x sin x 2 cos x C Such repeated...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
168 Chapter 8 Techniques of Integration At first this looks useless—we’re right back to R sec 3 x dx . But looking more closely: Z sec 3 x dx = sec x tan x - Z sec 3 x dx + Z sec x dx Z sec 3 x dx + Z sec 3 x dx = sec x tan x + Z sec x dx 2 Z sec 3 x dx = sec x tan x + Z sec x dx Z sec 3 x dx = sec x tan x 2 + 1 2 Z sec x dx = sec x tan x 2 + ln | sec x + tan x | 2 + C. EXAMPLE 8.14 Evaluate Z x 2 sin x dx . Let u = x 2 , dv = sin x dx ; then du = 2 x dx and v = - cos x . Now Z x 2 sin x dx = - x 2 cos x + Z 2 x cos x dx . This is better than the original integral, but we need to do integration by parts again. Let u = 2 x , dv = cos x dx ; then du = 2 and v = sin x , and Z x 2 sin x dx = - x 2 cos x + Z 2 x cos x dx = - x 2 cos x + 2 x sin x - Z 2 sin x dx = - x
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x + 2 x sin x + 2 cos x + C. Such repeated use of integration by parts is fairly common, but it can be a bit tedious to accomplish, and it is easy to make errors, especially sign errors involving the subtraction in the formula. There is a nice tabular method to accomplish the calculation that minimizes the chance for error and speeds up the whole process. We illustrate with the previous example. Here is the table:...
View Full Document

{[ snackBarMessage ]}

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