{"id":725,"date":"2025-06-20T17:07:50","date_gmt":"2025-06-20T17:07:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=725"},"modified":"2025-07-17T15:50:22","modified_gmt":"2025-07-17T15:50:22","slug":"other-strategies-for-integration-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/other-strategies-for-integration-learn-it-1\/","title":{"raw":"Other Strategies for Integration: Learn It 1","rendered":"Other Strategies for Integration: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find integrals efficiently using an integral table<\/li>\r\n \t<li>Use technology to solve integration problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Tables of Integrals<\/h2>\r\nYou've learned several integration methods, but there are additional tools that can help you tackle more complex integrals or verify your answers. One of the most useful resources is <strong>integration tables<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<div>\r\n<h3>integration tables<\/h3>\r\nPre-computed lists of integrals and their antiderivatives that you can reference to evaluate or check your work quickly. <em>You'll find these in many calculus textbooks, including the appendices of this one.<\/em>\r\n\r\n<\/div>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Integration tables can be incredibly helpful, but you need to use them wisely. They're great for:<\/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>Quick evaluation<\/strong> of integrals that match standard forms<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Checking your work<\/strong> after solving an integral manually<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Finding patterns<\/strong> that might help with similar problems<\/li>\r\n<\/ul>\r\nKeep in mind that two completely correct solutions can look very different, so don't panic if your answer doesn't match the table exactly\u2014they might still be equivalent. Here's a perfect example of how the same integral can have multiple correct forms:\r\n<ul>\r\n \t<li>\r\n<p class=\"whitespace-pre-wrap break-words\">Using trigonometric substitution with [latex]x=\\tan\\theta[\/latex]:<\/p>\r\n\r\n<center>[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)+C[\/latex]<\/center><\/li>\r\n \t<li>\r\n<p class=\"whitespace-pre-wrap break-words\">Using hyperbolic substitution with [latex]x=\\text{sinh}\\theta[\/latex]:<\/p>\r\n\r\n<center>[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}={\\text{sinh}}^{-1}x+C[\/latex]<\/center><\/li>\r\n<\/ul>\r\n<div id=\"fs-id1165041848129\" class=\"unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">These look completely different, but they're actually the same! We can prove algebraically that [latex]{\\text{sinh}}^{-1}x=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\r\nTwo antiderivatives are equivalent if their difference is just a constant. This makes sense because the derivative of a constant is zero, so [latex]\\frac{d}{dx}[F(x) + C_1] = \\frac{d}{dx}[G(x) + C_2][\/latex] when [latex]F(x) - G(x) = \\text{constant}[\/latex].\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042085684\" data-type=\"problem\">\r\n<p id=\"fs-id1165041973418\">Use the table formula<\/p>\r\n\r\n<div id=\"fs-id1165041758432\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex]<\/div>\r\n<p id=\"fs-id1165040771022\">to evaluate [latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1165041913720\" data-type=\"solution\">\r\n<p id=\"fs-id1165041766265\">If we look at integration tables, we see that several formulas contain expressions of the form [latex]\\sqrt{{a}^{2}-{u}^{2}}[\/latex]. This expression is actually similar to [latex]\\sqrt{16-{e}^{2x}}[\/latex], where [latex]a=4[\/latex] and [latex]u={e}^{x}[\/latex]. Keep in mind that we must also have [latex]du={e}^{x}[\/latex]. Multiplying the numerator and the denominator of the given integral by [latex]{e}^{x}[\/latex] should help to put this integral in a useful form. Thus, we now have<\/p>\r\n\r\n<div id=\"fs-id1165040752681\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx=\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041797345\">Substituting [latex]u={e}^{x}[\/latex] and [latex]du={e}^{x}[\/latex] produces [latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du[\/latex]. From the integration table (#88 in Appendix A),<\/p>\r\n\r\n<div id=\"fs-id1165042045994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041845825\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165040798665\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx}&amp; ={\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx}\\hfill &amp; &amp; &amp; \\text{Substitute}u={e}^{x}\\text{and}du={e}^{x}dx.\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{\\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du}\\hfill &amp; &amp; &amp; \\text{Apply the formula using}a=4.\\hfill \\\\ &amp; =-\\frac{\\sqrt{{4}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{4}+C\\hfill &amp; &amp; &amp; \\text{Substitute}u={e}^{x}.\\hfill \\\\ &amp; =-\\frac{\\sqrt{16-{e}^{2x}}}{u}-{\\sin}^{-1}\\left(\\frac{{e}^{x}}{4}\\right)+C.\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\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 example above.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6722645&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=CE2ryAVDgJ4&amp;video_target=tpm-plugin-248rwand-CE2ryAVDgJ4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.5.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"3.5.1\" here (opens in new window)<\/a>.<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find integrals efficiently using an integral table<\/li>\n<li>Use technology to solve integration problems<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Tables of Integrals<\/h2>\n<p>You&#8217;ve learned several integration methods, but there are additional tools that can help you tackle more complex integrals or verify your answers. One of the most useful resources is <strong>integration tables<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<div>\n<h3>integration tables<\/h3>\n<p>Pre-computed lists of integrals and their antiderivatives that you can reference to evaluate or check your work quickly. <em>You&#8217;ll find these in many calculus textbooks, including the appendices of this one.<\/em><\/p>\n<\/div>\n<\/section>\n<p class=\"whitespace-normal break-words\">Integration tables can be incredibly helpful, but you need to use them wisely. They&#8217;re great for:<\/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>Quick evaluation<\/strong> of integrals that match standard forms<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Checking your work<\/strong> after solving an integral manually<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Finding patterns<\/strong> that might help with similar problems<\/li>\n<\/ul>\n<p>Keep in mind that two completely correct solutions can look very different, so don&#8217;t panic if your answer doesn&#8217;t match the table exactly\u2014they might still be equivalent. Here&#8217;s a perfect example of how the same integral can have multiple correct forms:<\/p>\n<ul>\n<li>\n<p class=\"whitespace-pre-wrap break-words\">Using trigonometric substitution with [latex]x=\\tan\\theta[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)+C[\/latex]<\/div>\n<\/li>\n<li>\n<p class=\"whitespace-pre-wrap break-words\">Using hyperbolic substitution with [latex]x=\\text{sinh}\\theta[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}={\\text{sinh}}^{-1}x+C[\/latex]<\/div>\n<\/li>\n<\/ul>\n<div id=\"fs-id1165041848129\" class=\"unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">These look completely different, but they&#8217;re actually the same! We can prove algebraically that [latex]{\\text{sinh}}^{-1}x=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\n<p>Two antiderivatives are equivalent if their difference is just a constant. This makes sense because the derivative of a constant is zero, so [latex]\\frac{d}{dx}[F(x) + C_1] = \\frac{d}{dx}[G(x) + C_2][\/latex] when [latex]F(x) - G(x) = \\text{constant}[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042085684\" data-type=\"problem\">\n<p id=\"fs-id1165041973418\">Use the table formula<\/p>\n<div id=\"fs-id1165041758432\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex]<\/div>\n<p id=\"fs-id1165040771022\">to evaluate [latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041913720\" data-type=\"solution\">\n<p id=\"fs-id1165041766265\">If we look at integration tables, we see that several formulas contain expressions of the form [latex]\\sqrt{{a}^{2}-{u}^{2}}[\/latex]. This expression is actually similar to [latex]\\sqrt{16-{e}^{2x}}[\/latex], where [latex]a=4[\/latex] and [latex]u={e}^{x}[\/latex]. Keep in mind that we must also have [latex]du={e}^{x}[\/latex]. Multiplying the numerator and the denominator of the given integral by [latex]{e}^{x}[\/latex] should help to put this integral in a useful form. Thus, we now have<\/p>\n<div id=\"fs-id1165040752681\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx=\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041797345\">Substituting [latex]u={e}^{x}[\/latex] and [latex]du={e}^{x}[\/latex] produces [latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du[\/latex]. From the integration table (#88 in Appendix A),<\/p>\n<div id=\"fs-id1165042045994\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\\frac{\\sqrt{{a}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{a}+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041845825\">Thus,<\/p>\n<div id=\"fs-id1165040798665\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{x}}dx}& ={\\displaystyle\\int \\frac{\\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx}\\hfill & & & \\text{Substitute}u={e}^{x}\\text{and}du={e}^{x}dx.\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du}\\hfill & & & \\text{Apply the formula using}a=4.\\hfill \\\\ & =-\\frac{\\sqrt{{4}^{2}-{u}^{2}}}{u}-{\\sin}^{-1}\\frac{u}{4}+C\\hfill & & & \\text{Substitute}u={e}^{x}.\\hfill \\\\ & =-\\frac{\\sqrt{16-{e}^{2x}}}{u}-{\\sin}^{-1}\\left(\\frac{{e}^{x}}{4}\\right)+C.\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/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 example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6722645&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=CE2ryAVDgJ4&amp;video_target=tpm-plugin-248rwand-CE2ryAVDgJ4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.5.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;3.5.1&#8221; here (opens in new window)<\/a>.<\/section>\n","protected":false},"author":15,"menu_order":26,"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":"- Select Header -","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\/725"}],"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":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/725\/revisions"}],"predecessor-version":[{"id":1279,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/725\/revisions\/1279"}],"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\/725\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=725"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=725"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=725"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=725"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}