{"id":697,"date":"2025-06-20T17:04:56","date_gmt":"2025-06-20T17:04:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=697"},"modified":"2025-09-05T16:43:37","modified_gmt":"2025-09-05T16:43:37","slug":"trigonometric-integrals-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/trigonometric-integrals-fresh-take\/","title":{"raw":"Integration by Parts: Fresh Take","rendered":"Integration by Parts: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Recognize when to use integration by parts compared to other integration methods<\/li>\r\n \t<li>Use the integration by parts formula to solve indefinite integrals<\/li>\r\n \t<li>Apply integration by parts to evaluate definite integrals<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Integration-by-Parts<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Think of integration by parts as making a strategic trade. When you're faced with an integral like [latex]\\int x\\sin x\u00a0 dx[\/latex], you can't use substitution because there's no clear \"inside function\" and its derivative. Instead, you're going to swap this integral for a different one that's easier to handle.<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int u , dv = uv - \\int v\u00a0 du[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This formula is your trading mechanism. You're essentially saying: \"I'll take this product integral and exchange it for something I can actually solve.\"<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Strategy Tips:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Choose wisely<\/strong>: Your [latex]u[\/latex] should get simpler when differentiated, and your [latex]dv[\/latex] should be something you can actually integrate<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Don't add constants<\/strong>: When finding [latex]v[\/latex], skip the \"+C\" \u2013 you'll add it at the very end<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Check your work<\/strong>: If your new integral [latex]\\int v , du[\/latex] looks worse than what you started with, try switching your [latex]u[\/latex] and [latex]dv[\/latex] choices<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Watch <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this video for an introduction to integration by parts<\/a> and visit <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this website for examples of integration by parts<\/a>.<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1165040747250\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx[\/latex] using the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv={e}^{2x}dx[\/latex].<\/p>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1165042051378\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165040681938\">Find [latex]du[\/latex] and [latex]v[\/latex], and use the previous example as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1165040642767\" data-type=\"solution\">\r\n<p id=\"fs-id1165043352090\">[latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx=\\frac{1}{2}x{e}^{2x}-\\frac{1}{4}{e}^{2x}+C[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=305&amp;end=430&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts305to430_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2>How to Choose [latex]u[\/latex] and[latex] dv[\/latex]<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Choosing which part should be [latex]u[\/latex] and which should be [latex]dv[\/latex] doesn't have to feel like guesswork. Think of <strong>LIATE<\/strong> as your GPS for navigation\u2014it won't work for every single route, but it'll get you to your destination most of the time.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The LIATE Priority System:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>L<\/strong>ogarithmic functions (like [latex]\\ln x[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>I<\/strong>nverse trig functions (like [latex]\\arcsin x[\/latex], [latex]\\arctan x[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>A<\/strong>lgebraic functions (like [latex]x^2[\/latex], [latex]3x[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>T<\/strong>rigonometric functions (like [latex]\\sin x[\/latex], [latex]\\cos x[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>E<\/strong>xponential functions (like [latex]e^x[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>The Rule:<\/strong> Pick [latex]u[\/latex] to be whichever function type appears <strong>first<\/strong> in this list.<\/p>\r\n<p class=\"whitespace-normal break-words\">LIATE isn't arbitrary\u2014it's based on what's practical. Functions at the top of the list (L and I) are terrible to integrate but get simpler when you differentiate them. Functions at the bottom (T and E) are integration-friendly and stay manageable when differentiated. Algebraic functions (A) sit in the middle because they usually play nice either way.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Strategy Tips:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Sometimes you need to be creative<\/strong>: If your integral has only one function (like [latex]\\int \\ln x , dx[\/latex]), use the constant function 1 as your second part<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Watch for special cases<\/strong>: Sometimes breaking the LIATE rule makes sense (like choosing [latex]dv = te^{t^2}dt[\/latex] instead of just [latex]e^{t^2}dt[\/latex] when you can actually integrate the first one)<\/li>\r\n \t<li>Before committing to your choices, ask yourself these two critical questions:\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li>\"Will my [latex]u[\/latex] get simpler when I differentiate it?\"\r\nYou want the derivative [latex]du[\/latex] to be less complicated than your original [latex]u[\/latex]. For example, [latex]\\ln x[\/latex] becomes [latex]\\frac{1}{x}dx[\/latex] (simpler), while [latex]e^{x^2}[\/latex] becomes [latex]2xe^{x^2}dx[\/latex] (more complicated).<\/li>\r\n \t<li>\"Can I actually integrate my [latex]dv[\/latex]?\"\r\nThere's no point choosing [latex]dv = e^{x^2}dx[\/latex] if you can't find [latex]v[\/latex]. Make sure you can handle the integration step before moving forward.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165041899924\" data-type=\"problem\">\r\n<p id=\"fs-id1165041899926\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x\\text{ln}xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1165041864895\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041864902\">Use [latex]u=\\text{ln}x[\/latex] and [latex]dv=xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1165040747233\" data-type=\"solution\">\r\n<p id=\"fs-id1165043331164\">[latex]\\frac{1}{2}{x}^{2}\\text{ln}x-\\frac{1}{4}{x}^{2}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=558&amp;end=647&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts558to647_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042295958\" data-type=\"problem\">\r\n<p id=\"fs-id1165042295960\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}\\sin{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1165042231011\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165042231018\">This is similar to the previous example: Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1165042230965\" data-type=\"solution\">\r\n<p id=\"fs-id1165042407711\">[latex]\\text{-}{x}^{2}\\cos{x}+2x\\sin{x}+2\\cos{x}+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1419&amp;end=1575&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1419to1575_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Integration by Parts for Definite Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Working with definite integrals using integration by parts is like following the same recipe you already know, but with one additional finishing touch at the end. You still use the exact same formula and decision-making process\u2014you just need to evaluate your result at the upper and lower limits.<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int_a^b u , dv = uv\\big|_a^b - \\int_a^b v , du[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice that [latex]uv\\big|_a^b[\/latex] means you evaluate [latex]uv[\/latex] at the upper limit minus [latex]uv[\/latex] at the lower limit, just like any other definite integral.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem Solving Strategies:<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Choose [latex]u[\/latex] and [latex]dv[\/latex]<\/strong> using LIATE or the two-question test<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find [latex]du[\/latex] and [latex]v[\/latex]<\/strong> exactly as before<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Apply the formula<\/strong> with the limits intact<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate<\/strong> [latex]uv\\big|_a^b[\/latex] at both limits<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Solve the remaining integral<\/strong> [latex]\\int_a^b v , du[\/latex] (which might need substitution or other techniques)<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">You get a specific numerical answer instead of a family of functions plus C. No constant of integration needed!<\/p>\r\n<p class=\"whitespace-normal break-words\">Always step back and ask if your numerical answer makes sense. For area problems, is your result positive and reasonable compared to basic geometric shapes? For volume problems, does it seem plausible compared to simple cylinders or cones with similar dimensions?<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042299600\" data-type=\"problem\">\r\n<p id=\"fs-id1165042299602\">Evaluate [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}x\\cos{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558819\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558819\"]\r\n<div id=\"fs-id1165042299659\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165042299667\">Use the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv=\\cos{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n<div id=\"fs-id1165042299639\" data-type=\"solution\">\r\n<p id=\"fs-id1165042583725\">[latex]\\frac{\\pi }{2}-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1913&amp;end=2043&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1913to2043_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.1 Integration by Parts\" here (opens in new window)<\/a>.\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Recognize when to use integration by parts compared to other integration methods<\/li>\n<li>Use the integration by parts formula to solve indefinite integrals<\/li>\n<li>Apply integration by parts to evaluate definite integrals<\/li>\n<\/ul>\n<\/section>\n<h2>Integration-by-Parts<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Think of integration by parts as making a strategic trade. When you&#8217;re faced with an integral like [latex]\\int x\\sin x\u00a0 dx[\/latex], you can&#8217;t use substitution because there&#8217;s no clear &#8220;inside function&#8221; and its derivative. Instead, you&#8217;re going to swap this integral for a different one that&#8217;s easier to handle.<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int u , dv = uv - \\int v\u00a0 du[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This formula is your trading mechanism. You&#8217;re essentially saying: &#8220;I&#8217;ll take this product integral and exchange it for something I can actually solve.&#8221;<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Strategy Tips:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Choose wisely<\/strong>: Your [latex]u[\/latex] should get simpler when differentiated, and your [latex]dv[\/latex] should be something you can actually integrate<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Don&#8217;t add constants<\/strong>: When finding [latex]v[\/latex], skip the &#8220;+C&#8221; \u2013 you&#8217;ll add it at the very end<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Check your work<\/strong>: If your new integral [latex]\\int v , du[\/latex] looks worse than what you started with, try switching your [latex]u[\/latex] and [latex]dv[\/latex] choices<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Watch <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this video for an introduction to integration by parts<\/a> and visit <a href=\"http:\/\/www.sosmath.com\/calculus\/integration\/byparts\/byparts.html\" target=\"_blank\" rel=\"noopener\">this website for examples of integration by parts<\/a>.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1165040747250\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx[\/latex] using the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv={e}^{2x}dx[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042051378\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165040681938\">Find [latex]du[\/latex] and [latex]v[\/latex], and use the previous example as a guide.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040642767\" data-type=\"solution\">\n<p id=\"fs-id1165043352090\">[latex]{\\displaystyle\\int }^{\\text{ }}x{e}^{2x}dx=\\frac{1}{2}x{e}^{2x}-\\frac{1}{4}{e}^{2x}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=305&amp;end=430&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts305to430_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2>How to Choose [latex]u[\/latex] and[latex]dv[\/latex]<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Choosing which part should be [latex]u[\/latex] and which should be [latex]dv[\/latex] doesn&#8217;t have to feel like guesswork. Think of <strong>LIATE<\/strong> as your GPS for navigation\u2014it won&#8217;t work for every single route, but it&#8217;ll get you to your destination most of the time.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The LIATE Priority System:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>L<\/strong>ogarithmic functions (like [latex]\\ln x[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>I<\/strong>nverse trig functions (like [latex]\\arcsin x[\/latex], [latex]\\arctan x[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>A<\/strong>lgebraic functions (like [latex]x^2[\/latex], [latex]3x[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>T<\/strong>rigonometric functions (like [latex]\\sin x[\/latex], [latex]\\cos x[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\"><strong>E<\/strong>xponential functions (like [latex]e^x[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>The Rule:<\/strong> Pick [latex]u[\/latex] to be whichever function type appears <strong>first<\/strong> in this list.<\/p>\n<p class=\"whitespace-normal break-words\">LIATE isn&#8217;t arbitrary\u2014it&#8217;s based on what&#8217;s practical. Functions at the top of the list (L and I) are terrible to integrate but get simpler when you differentiate them. Functions at the bottom (T and E) are integration-friendly and stay manageable when differentiated. Algebraic functions (A) sit in the middle because they usually play nice either way.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Strategy Tips:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Sometimes you need to be creative<\/strong>: If your integral has only one function (like [latex]\\int \\ln x , dx[\/latex]), use the constant function 1 as your second part<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Watch for special cases<\/strong>: Sometimes breaking the LIATE rule makes sense (like choosing [latex]dv = te^{t^2}dt[\/latex] instead of just [latex]e^{t^2}dt[\/latex] when you can actually integrate the first one)<\/li>\n<li>Before committing to your choices, ask yourself these two critical questions:\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li>&#8220;Will my [latex]u[\/latex] get simpler when I differentiate it?&#8221;<br \/>\nYou want the derivative [latex]du[\/latex] to be less complicated than your original [latex]u[\/latex]. For example, [latex]\\ln x[\/latex] becomes [latex]\\frac{1}{x}dx[\/latex] (simpler), while [latex]e^{x^2}[\/latex] becomes [latex]2xe^{x^2}dx[\/latex] (more complicated).<\/li>\n<li>&#8220;Can I actually integrate my [latex]dv[\/latex]?&#8221;<br \/>\nThere&#8217;s no point choosing [latex]dv = e^{x^2}dx[\/latex] if you can&#8217;t find [latex]v[\/latex]. Make sure you can handle the integration step before moving forward.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165041899924\" data-type=\"problem\">\n<p id=\"fs-id1165041899926\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}x\\text{ln}xdx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Hint<\/button><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041864895\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041864902\">Use [latex]u=\\text{ln}x[\/latex] and [latex]dv=xdx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Show Solution<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040747233\" data-type=\"solution\">\n<p id=\"fs-id1165043331164\">[latex]\\frac{1}{2}{x}^{2}\\text{ln}x-\\frac{1}{4}{x}^{2}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=558&amp;end=647&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts558to647_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042295958\" data-type=\"problem\">\n<p id=\"fs-id1165042295960\">Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}\\sin{x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558889\">Hint<\/button><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042231011\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165042231018\">This is similar to the previous example: Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{2}{e}^{3x}dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558890\">Show Solution<\/button><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042230965\" data-type=\"solution\">\n<p id=\"fs-id1165042407711\">[latex]\\text{-}{x}^{2}\\cos{x}+2x\\sin{x}+2\\cos{x}+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1419&amp;end=1575&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1419to1575_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2 data-type=\"title\">Integration by Parts for Definite Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Working with definite integrals using integration by parts is like following the same recipe you already know, but with one additional finishing touch at the end. You still use the exact same formula and decision-making process\u2014you just need to evaluate your result at the upper and lower limits.<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int_a^b u , dv = uv\\big|_a^b - \\int_a^b v , du[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Notice that [latex]uv\\big|_a^b[\/latex] means you evaluate [latex]uv[\/latex] at the upper limit minus [latex]uv[\/latex] at the lower limit, just like any other definite integral.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem Solving Strategies:<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Choose [latex]u[\/latex] and [latex]dv[\/latex]<\/strong> using LIATE or the two-question test<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find [latex]du[\/latex] and [latex]v[\/latex]<\/strong> exactly as before<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Apply the formula<\/strong> with the limits intact<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Evaluate<\/strong> [latex]uv\\big|_a^b[\/latex] at both limits<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Solve the remaining integral<\/strong> [latex]\\int_a^b v , du[\/latex] (which might need substitution or other techniques)<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">You get a specific numerical answer instead of a family of functions plus C. No constant of integration needed!<\/p>\n<p class=\"whitespace-normal break-words\">Always step back and ask if your numerical answer makes sense. For area problems, is your result positive and reasonable compared to basic geometric shapes? For volume problems, does it seem plausible compared to simple cylinders or cones with similar dimensions?<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042299600\" data-type=\"problem\">\n<p id=\"fs-id1165042299602\">Evaluate [latex]{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}x\\cos{x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558819\">Hint<\/button><\/p>\n<div id=\"q44558819\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042299659\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165042299667\">Use the integration-by-parts formula with [latex]u=x[\/latex] and [latex]dv=\\cos{x}dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558829\">Show Solution<\/button><\/p>\n<div id=\"q44558829\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042299639\" data-type=\"solution\">\n<p id=\"fs-id1165042583725\">[latex]\\frac{\\pi }{2}-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/lpmf-HAmaEU?controls=0&amp;start=1913&amp;end=2043&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.1IntegrationByParts1913to2043_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.1 Integration by Parts&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":11,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":541,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/697"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/697\/revisions"}],"predecessor-version":[{"id":2224,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/697\/revisions\/2224"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/541"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/697\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=697"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=697"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=697"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=697"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}