{"id":17092,"date":"2016-07-30T17:49:22","date_gmt":"2016-07-30T17:49:22","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=17092"},"modified":"2016-07-30T17:49:22","modified_gmt":"2016-07-30T17:49:22","slug":"calculus-integration-by-parts-%e2%88%abexsinxdx","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/calculus-integration-by-parts-%e2%88%abexsinxdx\/","title":{"rendered":"Calculus: integration by parts: \u222bex<\/sup>sinxdx"},"content":{"rendered":"

The tutor shows how to integrate ex<\/sup>sinxdx by parts.<\/h1>\n

I’ve written a couple of posts on integration by parts: here<\/a> and here<\/a>. The method is used on products, and depends on choosing one function to integrate, then differentiating the other:<\/p>\n

\u222buv’=uv-\u222bvu’<\/p>\n

With one of the functions xk<\/sup> for some integer k, that one will typically be differentiated, since it will eventually disappear (after k iterations). For k>2, the process will perhaps be long and messy, with lots of chances for error, but likely possible nonetheless – especially if the other function is integrable.<\/p>\n

What if neither function will disappear, no matter how many times it’s differentiated? Today’s example is such a case:<\/p>\n

\u222bex<\/sup>sinxdx<\/p>\n

Happily, ex<\/sup> is easily integrable. Let’s imagine it as \u222b(sinx)ex<\/sup>dx: we’ll integrate ex<\/sup>dx to ex<\/sup>, then differentiate sinx to cosx:<\/p>\n

\u222buv’=uv-\u222bvu’<\/p>\n

becomes<\/p>\n

\u222b(sinx)ex<\/sup>dx=(sinx)ex<\/sup> -\u222bex<\/sup>cosxdx<\/p>\n

For the second round, we’ll integrate ex<\/sup>dx again, while differentiating cosx:<\/p>\n

\u222bex<\/sup>cosxdx=(cosx)ex<\/sup> – \u222bex<\/sup>(-sinx)dx<\/p>\n

⇒\u222bex<\/sup>cosxdx=(cosx)ex<\/sup> + \u222bex<\/sup>(sinx)dx<\/p>\n

Notice that our original integral, \u222bex<\/sup>sinxdx, has emerged at the end of the second iteration. Believe it or not, this is exactly what we want. Let \u222bex<\/sup>sinxdx = A. Substituting and carefully retracing the procedure, we arrive at<\/p>\n

A = ex<\/sup>sinx – [ex<\/sup>cosx + A]<\/p>\n

or<\/p>\n

A = ex<\/sup>sinx – ex<\/sup>cosx – A<\/p>\n

Adding A to both sides we get<\/p>\n

2A = ex<\/sup>sinx – ex<\/sup>cosx <\/p>\n

Dividing both sides by 2, then resubstituting \u222bex<\/sup>sinxdx for A, gives<\/p>\n

A = \u222bex<\/sup>sinxdx = (ex<\/sup>sinx – ex<\/sup>cosx)\/2 + C <\/p>\n

I’ll be talking more about integral calculus in coming posts:)<\/p>\n

Source:<\/p>\n

Larson, Roland and Robert Hostetler. Calculus<\/u>. Toronto: D C Heath and Company, 1989.<\/p>\n

Jack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"

The tutor shows how to integrate exsinxdx by parts. I’ve written a couple of posts on integration by parts: here and here. The method is used on products, and depends on choosing one function to integrate, then differentiating the other: …<\/p>\n

Calculus: integration by parts: \u222bex<\/sup>sinxdx<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[234,3],"tags":[1750,1759,1760,1758],"class_list":["post-17092","post","type-post","status-publish","format-standard","hentry","category-calculus","category-math","tag-integration-by-parts","tag-integration-by-parts-with-substitution","tag-recursive-integration-by-parts","tag-repeated-integration-by-parts"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17092","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=17092"}],"version-history":[{"count":41,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17092\/revisions"}],"predecessor-version":[{"id":17133,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/17092\/revisions\/17133"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=17092"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=17092"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=17092"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}