{"id":1217,"date":"2025-07-16T15:17:04","date_gmt":"2025-07-16T15:17:04","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1217"},"modified":"2025-09-10T14:26:15","modified_gmt":"2025-09-10T14:26:15","slug":"trigonometric-integrals-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/trigonometric-integrals-learn-it-5\/","title":{"raw":"Trigonometric Integrals: Learn It 5","rendered":"Trigonometric Integrals: Learn It 5"},"content":{"raw":"

Integrating Expressions Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h2>\r\nNow let's tackle expressions involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]. This requires a different approach than what we just learned.\r\n\r\nThe key difference here is domain<\/strong>. Unlike [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], the expression [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex] is defined for all real values<\/strong> of [latex]x[\/latex]. This means we need a trigonometric function that also has a range of all real numbers.\r\n

Our options:<\/strong><\/p>\r\n\r\n

    \r\n \t
  • [latex]x = a\\tan\\theta[\/latex] \u2713 (tangent has range [latex](-\\infty, \\infty)[\/latex])<\/li>\r\n \t
  • [latex]x = a\\cot\\theta[\/latex] \u2713 (cotangent has range [latex](-\\infty, \\infty)[\/latex])<\/li>\r\n<\/ul>\r\n

    Either of these substitutions would actually work, but the standard substitution is [latex]x=a\\tan\\theta [\/latex] or, equivalently, [latex]\\tan\\theta =\\frac{x}{a}[\/latex], where [latex]-\\frac{\\pi}{2} < \\theta < \\frac{\\pi}{2}[\/latex].<\/p>\r\n\r\n

    Why This Substitution Works<\/h3>\r\n

    When you substitute [latex]x = a\\tan\\theta[\/latex]:<\/p>\r\n

    [latex]\\sqrt{{a}^{2}+{x}^{2}} = \\sqrt{{a}^{2}+{(a\\tan\\theta)}^{2}} = \\sqrt{{a}^{2}(1+{\\tan}^{2}\\theta)} = \\sqrt{{a}^{2}{\\sec}^{2}\\theta} = a\\sec\\theta[\/latex]<\/p>\r\n\r\n

    This substitution relies on the identity [latex]1 + \\tan^2\\theta = \\sec^2\\theta[\/latex]. Since [latex]\\sec\\theta > 0[\/latex] over the interval [latex](-\\frac{\\pi}{2}, \\frac{\\pi}{2})[\/latex], we have [latex]|a\\sec\\theta| = a\\sec\\theta[\/latex].<\/section>\r\n

    Since [latex]\\tan\\theta = \\frac{x}{a}[\/latex], you can build this reference triangle:<\/p>\r\n\r\n

      \r\n \t
    • Adjacent side: [latex]a[\/latex]<\/li>\r\n \t
    • Opposite side: [latex]x[\/latex]<\/li>\r\n \t
    • Hypotenuse: [latex]\\sqrt{a^2+x^2}[\/latex] (by Pythagorean theorem)<\/li>\r\n \t
    • Therefore: [latex]\\theta = \\tan^{-1}\\left(\\frac{x}{a}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This A reference triangle can be constructed to express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].[\/caption]
      \"Caution\"The reference triangle assumes [latex]x > 0[\/latex], but don't worry! The trigonometric ratios it produces work for all values of [latex]x[\/latex], including when [latex]x \\leq 0[\/latex].<\/section>
      Problem-Solving Strategy for [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/strong>\r\n
        \r\n \t
      1. Check first:<\/strong> Can this integral be solved more easily another way? In some cases, it is more convenient to use an alternative method.<\/li>\r\n \t
      2. Substitute:<\/strong> [latex]x = a\\tan\\theta[\/latex] and [latex]dx = a\\sec^2\\theta , d\\theta[\/latex].<\/li>\r\n \t
      3. Simplify:<\/strong> Use [latex]\\sqrt{{a}^{2}+{x}^{2}} = a\\sec\\theta[\/latex]<\/li>\r\n \t
      4. Integrate:<\/strong> Use trigonometric integration techniques<\/li>\r\n \t
      5. Convert back:<\/strong> Use the reference triangle to express the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\tan}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
        \r\n
        \r\n

        Evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex] and check the solution by differentiating.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n

        \r\n

        Begin with the substitution [latex]x=\\tan\\theta [\/latex] and [latex]dx={\\sec}^{2}\\theta d\\theta [\/latex]. Since [latex]\\tan\\theta =x[\/latex], draw the reference triangle in the following figure.<\/p>\r\n\r\n

        <\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 5. The reference triangle for this example.[\/caption]<\/figure>\r\n

        Thus,<\/p>\r\n\r\n

        [latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}}& ={\\displaystyle\\int \\frac{{\\sec}^{2}\\theta }{\\sec\\theta }d\\theta }\\hfill & & & \\begin{array}{c}\\text{Substitute}x=\\tan\\theta \\text{and}dx={\\sec}^{2}\\theta d\\theta .\\text{This}\\hfill \\\\ \\text{substitution makes}\\sqrt{1+{x}^{2}}=\\sec\\theta .\\text{Simplify.}\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int ^{\\text{ }}\\sec\\theta d\\theta }\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =\\text{ln}|\\sec\\theta +\\tan\\theta |+C\\hfill & & & \\begin{array}{c}\\text{Use the reference triangle to express the result}\\hfill \\\\ \\text{in terms of}x.\\hfill \\end{array}\\hfill \\\\ & =\\text{ln}|\\sqrt{1+{x}^{2}}+x|+C.\\hfill & & & \\end{array}[\/latex]<\/div>\r\n \r\n

        To check the solution, differentiate:<\/p>\r\n\r\n

        [latex]\\begin{array}{cc}\\hfill \\frac{d}{dx}\\left(\\text{ln}|\\sqrt{1+{x}^{2}}+x|\\right)& =\\frac{1}{\\sqrt{1+{x}^{2}}+x}\\cdot \\left(\\frac{x}{\\sqrt{1+{x}^{2}}}+1\\right)\\hfill \\\\ & =\\frac{1}{\\sqrt{1+{x}^{2}}+x}\\cdot \\frac{x+\\sqrt{1+{x}^{2}}}{\\sqrt{1+{x}^{2}}}\\hfill \\\\ & =\\frac{1}{\\sqrt{1+{x}^{2}}}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

        Since [latex]\\sqrt{1+{x}^{2}}+x>0[\/latex] for all values of [latex]x[\/latex], we could rewrite [latex]\\text{ln}|\\sqrt{1+{x}^{2}}+x|+C=\\text{ln}\\left(\\sqrt{1+{x}^{2}}+x\\right)+C[\/latex], if desired.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>In the example below, we explore how to use hyperbolic trigonometric functions as an alternative substitution method. First, we briefly review the definition of these functions and their derivatives.\r\n\r\n

        For any real number [latex] x [\/latex], the hyperbolic sine and hyperbolic cosine are defined as:\r\n

        [latex] \\sinh x = \\frac{e^x - e^{-x}}{2} \\: \\text{and}\\:\\cosh x = \\frac{e^x + e^{-x}}{2} [\/latex]<\/p>\r\nTheir derivatives are given by:\r\n

        [latex] \\frac{d}{dx} \\left( \\sinh x \\right) = \\cosh x \\:\\text{and}\\:\\frac{d}{dx} \\left( \\cosh x \\right) = \\sinh x [\/latex]<\/p>\r\n\r\n<\/section>

        \r\n
        \r\n

        Use the substitution [latex]x=\\text{sinh}\\theta [\/latex] to evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n

        \r\n

        Because [latex]\\text{sinh}\\theta [\/latex] has a range of all real numbers, and [latex]1+{\\text{sinh}}^{2}\\theta ={\\text{cosh}}^{2}\\theta [\/latex], we may also use the substitution [latex]x=\\text{sinh}\\theta [\/latex] to evaluate this integral. In this case, [latex]dx=\\text{cosh}\\theta d\\theta [\/latex]. Consequently,<\/p>\r\n\r\n

        [latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}}& ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\sqrt{1+{\\text{sinh}}^{2}\\theta }}d\\theta }\\hfill & & & \\begin{array}{c}\\text{Substitute}x=\\text{sinh}\\theta \\text{and}dx=\\text{cosh}\\theta d\\theta .\\hfill \\\\ \\text{Substitute}1+{\\text{sinh}}^{2}\\theta ={\\text{cosh}}^{2}\\theta .\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\sqrt{{\\text{cosh}}^{2}\\theta }}d\\theta }\\hfill & & & \\sqrt{{\\text{cosh}}^{2}\\theta }=|\\text{cosh}\\theta |\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{|\\text{cosh}\\theta |}d\\theta }\\hfill & & & |\\text{cosh}\\theta |=\\text{cosh}\\theta \\text{since}\\text{cosh}\\theta >0\\text{for all}\\theta .\\hfill \\\\ & ={\\displaystyle\\int \\frac{\\text{cosh}\\theta }{\\text{cosh}\\theta }d\\theta }\\hfill & & & \\text{Simplify.}\\hfill \\\\ & ={\\displaystyle\\int ^{\\text{ }}1d\\theta }\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =\\theta +C\\hfill & & & \\text{Since}x=\\text{sinh}\\theta ,\\text{we know}\\theta ={\\text{sinh}}^{-1}x.\\hfill \\\\ & ={\\text{sinh}}^{-1}x+C.\\hfill & & & \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n
        \r\n
        Analysis<\/strong><\/div>\r\n

        This answer looks quite different from the answer obtained using the substitution [latex]x=\\tan\\theta [\/latex]. To see that the solutions are the same, set [latex]y={\\text{sinh}}^{-1}x[\/latex]. Thus, [latex]\\text{sinh}y=x[\/latex]. From this equation we obtain:<\/p>\r\n\r\n

        [latex]\\frac{{e}^{y}-{e}^{\\text{-}y}}{2}=x[\/latex].<\/div>\r\n \r\n

        After multiplying both sides by [latex]2{e}^{y}[\/latex] and rewriting, this equation becomes:<\/p>\r\n\r\n

        [latex]{e}^{2y}-2x{e}^{y}-1=0[\/latex].<\/div>\r\n \r\n

        Use the quadratic equation to solve for [latex]{e}^{y}\\text{:}[\/latex]<\/p>\r\n\r\n

        [latex]{e}^{y}=\\frac{2x\\pm \\sqrt{4{x}^{2}+4}}{2}[\/latex].<\/div>\r\n \r\n

        Simplifying, we have:<\/p>\r\n\r\n

        [latex]{e}^{y}=x\\pm \\sqrt{{x}^{2}+1}[\/latex].<\/div>\r\n \r\n

        Since [latex]x-\\sqrt{{x}^{2}+1}<0[\/latex], it must be the case that [latex]{e}^{y}=x+\\sqrt{{x}^{2}+1}[\/latex]. Thus,<\/p>\r\n\r\n

        [latex]y=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\r\n \r\n

        Last, we obtain<\/p>\r\n\r\n

        [latex]{\\text{sinh}}^{-1}x=\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)[\/latex].<\/div>\r\n \r\n

        After we make the final observation that, since [latex]x+\\sqrt{{x}^{2}+1}>0[\/latex],<\/p>\r\n\r\n

        [latex]\\text{ln}\\left(x+\\sqrt{{x}^{2}+1}\\right)=\\text{ln}|\\sqrt{1+{x}^{2}}+x|[\/latex],<\/div>\r\n \r\n

        we see that the two different methods produced equivalent solutions.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

        [ohm_question hide_question_numbers=1]311296[\/ohm_question]<\/section>","rendered":"

        Integrating Expressions Involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/h2>\n

        Now let’s tackle expressions involving [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]. This requires a different approach than what we just learned.<\/p>\n

        The key difference here is domain<\/strong>. Unlike [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex], the expression [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex] is defined for all real values<\/strong> of [latex]x[\/latex]. This means we need a trigonometric function that also has a range of all real numbers.<\/p>\n

        Our options:<\/strong><\/p>\n

          \n
        • [latex]x = a\\tan\\theta[\/latex] \u2713 (tangent has range [latex](-\\infty, \\infty)[\/latex])<\/li>\n
        • [latex]x = a\\cot\\theta[\/latex] \u2713 (cotangent has range [latex](-\\infty, \\infty)[\/latex])<\/li>\n<\/ul>\n

          Either of these substitutions would actually work, but the standard substitution is [latex]x=a\\tan\\theta[\/latex] or, equivalently, [latex]\\tan\\theta =\\frac{x}{a}[\/latex], where [latex]-\\frac{\\pi}{2} < \\theta < \\frac{\\pi}{2}[\/latex].<\/p>\n

          Why This Substitution Works<\/h3>\n

          When you substitute [latex]x = a\\tan\\theta[\/latex]:<\/p>\n

          [latex]\\sqrt{{a}^{2}+{x}^{2}} = \\sqrt{{a}^{2}+{(a\\tan\\theta)}^{2}} = \\sqrt{{a}^{2}(1+{\\tan}^{2}\\theta)} = \\sqrt{{a}^{2}{\\sec}^{2}\\theta} = a\\sec\\theta[\/latex]<\/p>\n

          This substitution relies on the identity [latex]1 + \\tan^2\\theta = \\sec^2\\theta[\/latex]. Since [latex]\\sec\\theta > 0[\/latex] over the interval [latex](-\\frac{\\pi}{2}, \\frac{\\pi}{2})[\/latex], we have [latex]|a\\sec\\theta| = a\\sec\\theta[\/latex].<\/section>\n

          Since [latex]\\tan\\theta = \\frac{x}{a}[\/latex], you can build this reference triangle:<\/p>\n

            \n
          • Adjacent side: [latex]a[\/latex]<\/li>\n
          • Opposite side: [latex]x[\/latex]<\/li>\n
          • Hypotenuse: [latex]\\sqrt{a^2+x^2}[\/latex] (by Pythagorean theorem)<\/li>\n
          • Therefore: [latex]\\theta = \\tan^{-1}\\left(\\frac{x}{a}\\right)[\/latex]<\/li>\n<\/ul>\n
            \"This
            A reference triangle can be constructed to express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].<\/figcaption><\/figure>\n
            \"Caution\"The reference triangle assumes [latex]x > 0[\/latex], but don’t worry! The trigonometric ratios it produces work for all values of [latex]x[\/latex], including when [latex]x \\leq 0[\/latex].<\/section>\n
            Problem-Solving Strategy for [latex]\\sqrt{{a}^{2}+{x}^{2}}[\/latex]<\/strong><\/p>\n
              \n
            1. Check first:<\/strong> Can this integral be solved more easily another way? In some cases, it is more convenient to use an alternative method.<\/li>\n
            2. Substitute:<\/strong> [latex]x = a\\tan\\theta[\/latex] and [latex]dx = a\\sec^2\\theta , d\\theta[\/latex].<\/li>\n
            3. Simplify:<\/strong> Use [latex]\\sqrt{{a}^{2}+{x}^{2}} = a\\sec\\theta[\/latex]<\/li>\n
            4. Integrate:<\/strong> Use trigonometric integration techniques<\/li>\n
            5. Convert back:<\/strong> Use the reference triangle to express the result in terms of [latex]x[\/latex]. You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\tan}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].<\/li>\n<\/ol>\n<\/section>\n
              \n
              \n

              Evaluate [latex]\\displaystyle\\int \\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex] and check the solution by differentiating.<\/p>\n<\/div>\n