Sometimes, a derivative is done using the chain rule, and it leaves with an equation that, at first glance, can look intractable when we are attempting to integrate it. In those cases, although the problem may look difficult or often will look like something that can only be solved through the use of Integration by Parts, there often exists a much simpler solution: that of U-Substitution.<\/p>\n
Suppose we have an equation in the form \u222bu\u2019(x)f(u(x))dx. Then we can do a substitution in order to make the equation easier to integrate in a way that we will see shortly.
\nAlthough the way it was phrased above may make it sound intimidating, in reality the process is very straightforward.<\/p>\n
Suppose we wish to find \u222bx3<\/sup>sin(x4<\/sup>)dx Although this may look tricky, once we notice that the derivative of x4<\/sup> includes x3<\/sup>, it becomes much more straightforward. \u222bsin(u)du\/4 = 1\/4 \u222bsin(u)du If all of the xs have not been eliminated, then we cannot go through with the substitution. A good choice of u will leave no x or dx in the equation at all, only u and du.<\/p>\n Interested in calculus tutoring services<\/a>? Learn more about how we are assisting thousands of students each academic year.<\/p>\n SchoolTutoring Academy<\/span> is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in Cumberland, MD: visit Tutoring in Cumberland, MD<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":" Overview Sometimes, a derivative is done using the chain rule, and it leaves with an equation that, at first glance, can look intractable when we are attempting to integrate it. In those cases, although the problem may look difficult or often will look like something that can only be solved through the use of Integration […]<\/p>\n","protected":false},"author":29,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[4],"tags":[2789,2790],"class_list":["post-7344","post","type-post","status-publish","format-standard","hentry","category-calculus","tag-solving-integrals","tag-u-substitution"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/7344","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/users\/29"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/comments?post=7344"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/7344\/revisions"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/media?parent=7344"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/categories?post=7344"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/tags?post=7344"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nWe let u=x4<\/sup>
\nThen in that case, du\/dx=4x3<\/sup>, or du\/4 = x3<\/sup>dx
\nIf we move things around in the initial integral, we can note that
\n\u222bx3<\/sup>sin(x4<\/sup>)dxdx = \u222bsin(x4<\/sup>)x3<\/sup>dx
\nNow, since x4<\/sup> = u, and x3<\/sup>dx = du\/4, we can substitute in for the x as such<\/p>\n
\nWhich is \u00bc(-cos(u))+C = -\u00bccos(u)+C
\nAnd now we need to substitute back for u, so
\n-\u00bccos(u)+C = -\u00bccos(x4<\/sup>)+C<\/p>\nCommon Mistakes<\/h3>\n