{"id":702,"date":"2025-06-20T17:06:12","date_gmt":"2025-06-20T17:06:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=702"},"modified":"2025-09-10T14:12:53","modified_gmt":"2025-09-10T14:12:53","slug":"trigonometric-integrals-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/trigonometric-integrals-learn-it-2\/","title":{"raw":"Trigonometric Integrals: Learn It 2","rendered":"Trigonometric Integrals: Learn It 2"},"content":{"raw":"

Integrating Products and Powers of [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex]<\/h2>\r\nBefore discussing the integration of products and powers of [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex], it is useful to recall the integrals involving [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex] we have already learned:\r\n\r\n
\r\n
    \r\n \t
  1. [latex]{\\displaystyle\\int}{\\sec}^{2}xdx=\\tan{x}+C[\/latex]<\/li>\r\n \t
  2. [latex]{\\displaystyle\\int}\\sec{x}\\tan{x}dx=\\sec{x}+C[\/latex]<\/li>\r\n \t
  3. [latex]{\\displaystyle\\int}\\tan{x}dx=\\text{ln}|\\sec{x}|+C[\/latex]<\/li>\r\n \t
  4. [latex]{\\displaystyle\\int}\\sec{x}dx=\\text{ln}|\\sec{x}+\\tan{x}|+C[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>For most integrals of products and powers of [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex], we rewrite the expression we wish to integrate as the sum or difference of integrals of the form [latex]{\\displaystyle\\int}{\\tan}^{j}x{\\sec}^{2}xdx[\/latex] or [latex]{\\displaystyle\\int}{\\sec}^{j}x\\tan{x}dx[\/latex]. As we see in the following example, we can evaluate these new integrals by using u<\/em>-substitution.\u00a0 Before doing so, it is useful to note how the Pythagorean Identity implies relationships between other pairs of trigonometric functions.\r\n\r\n
    \r\n

    For any angle [latex] x [\/latex]:<\/p>\r\n\r\n

    [latex] \\sin^2 x + \\cos^2 x = 1 [\/latex]<\/center>Dividing the original equation by [latex] \\cos^2 x [\/latex] and simplifying yields an expression for [latex] \\sec^2 x [\/latex] in terms of [latex] \\tan^2 x [\/latex]:\r\n\r\n
    [latex] \\tan^2 x + 1 = \\sec^2 x [\/latex]<\/center>Subtracting both sides of the equation by [latex] 1 [\/latex] yields an expression for [latex] \\tan^2 x [\/latex] in terms of [latex] \\sec^2 x [\/latex]:\r\n\r\n
    [latex] \\tan^2 x = \\sec^2 x - 1 [\/latex]<\/center><\/section>
    \r\n
    \r\n

    Evaluate [latex]{\\displaystyle\\int}{\\sec}^{5}x\\tan{x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558499\"]\r\n

    \r\n

    Start by rewriting [latex]{\\sec}^{5}x\\tan{x}[\/latex] as [latex]{\\sec}^{4}x\\sec{x}\\tan{x}[\/latex].<\/p>\r\n\r\n

    [latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int}{\\sec}^{5}x\\tan{x}dx& ={\\displaystyle\\int}{\\sec}^{4}x\\sec{x}\\tan{x}dx\\hfill & & \\text{Let }u=\\sec{x};\\text{then},du=\\sec{x}\\tan{x}dx.\\hfill \\\\ & ={\\displaystyle\\int}{u}^{4}du\\hfill & & \\text{Evaluate the integral}.\\hfill \\\\ & =\\frac{1}{5}{u}^{5}+C\\hfill & & \\text{Substitute }\\sec{x}=u.\\hfill \\\\ & =\\frac{1}{5}{\\sec}^{5}x+C\\hfill & & \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
    \r\n
    \r\n

    Evaluate [latex]{\\displaystyle\\int}{\\tan}^{5}x{\\sec}^{2}xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558299\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558299\"]\r\n

    \r\n

    Let [latex]u=\\tan{x}[\/latex] and [latex]du={\\sec}^{2}x[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558399\"]\r\n

    \r\n

    [latex]\\frac{1}{6}{\\tan}^{6}x+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>We now take a look at the various strategies for integrating products and powers of [latex]\\sec{x}[\/latex] and [latex]\\tan{x}[\/latex].\r\n\r\n

    Problem-Solving Strategy: Integrating [latex]{\\displaystyle\\int}{\\tan}^{k}x{\\sec}^{j}xdx[\/latex]<\/strong>\r\n
      \r\n \t
    1. If [latex]j[\/latex] is even and [latex]j\\ge 2[\/latex], rewrite [latex]{\\sec}^{j}x={\\sec}^{j - 2}x{\\sec}^{2}x[\/latex] and use [latex]{\\sec}^{2}x={\\tan}^{2}x+1[\/latex] to rewrite [latex]{\\sec}^{j - 2}x[\/latex] in terms of [latex]\\tan{x}[\/latex]. Let [latex]u=\\tan{x}[\/latex] and [latex]du={\\sec}^{2}x[\/latex].<\/li>\r\n \t
    2. If [latex]k[\/latex] is odd and [latex]j\\ge 1[\/latex], rewrite [latex]{\\tan}^{k}x{\\sec}^{j}x={\\tan}^{k - 1}x{\\sec}^{j - 1}x\\sec{x}\\tan{x}[\/latex] and use [latex]{\\tan}^{2}x={\\sec}^{2}x - 1[\/latex] to rewrite [latex]{\\tan}^{k - 1}x[\/latex] in terms of [latex]\\sec{x}[\/latex]. Let [latex]u=\\sec{x}[\/latex] and [latex]du=\\sec{x}\\tan{x}dx[\/latex]. (Note<\/em>: If [latex]j[\/latex] is even and [latex]k[\/latex] is odd, then either strategy 1 or strategy 2 may be used.)<\/li>\r\n \t
    3. If [latex]k[\/latex] is odd where [latex]k\\ge 3[\/latex] and [latex]j=0[\/latex], rewrite [latex]{\\tan}^{k}x={\\tan}^{k - 2}x{\\tan}^{2}x={\\tan}^{k - 2}x\\left({\\sec}^{2}x - 1\\right)={\\tan}^{k - 2}x{\\sec}^{2}x-{\\tan}^{k - 2}x[\/latex]. It may be necessary to repeat this process on the [latex]{\\tan}^{k - 2}x[\/latex] term.<\/li>\r\n \t
    4. If [latex]k[\/latex] is even and [latex]j[\/latex] is odd, then use [latex]{\\tan}^{2}x={\\sec}^{2}x - 1[\/latex] to express [latex]{\\tan}^{k}x[\/latex] in terms of [latex]\\sec{x}[\/latex]. Use integration by parts to integrate odd powers of [latex]\\sec{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
      \r\n
      \r\n

      Evaluate [latex]{\\displaystyle\\int}{\\tan}^{6}x{\\sec}^{4}xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558199\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558199\"]\r\n

      \r\n

      Since the power on [latex]\\sec{x}[\/latex] is even, rewrite [latex]{\\sec}^{4}x={\\sec}^{2}x{\\sec}^{2}x[\/latex] and use [latex]{\\sec}^{2}x={\\tan}^{2}x+1[\/latex] to rewrite the first [latex]{\\sec}^{2}x[\/latex] in terms of [latex]\\tan{x}[\/latex]. Thus,<\/p>\r\n\r\n

      [latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int}{\\tan}^{6}x{\\sec}^{4}xdx& ={\\displaystyle\\int}{\\tan}^{6}x\\left({\\tan}^{2}x+1\\right){\\sec}^{2}xdx\\hfill & & \\text{Let }u=\\tan{x}\\text{ and }du={\\sec}^{2}x.\\hfill \\\\ & ={\\displaystyle\\int}{u}^{6}\\left({u}^{2}+1\\right)du\\hfill & & \\text{Expand}.\\hfill \\\\ & ={\\displaystyle\\int}\\left({u}^{8}+{u}^{6}\\right)du\\hfill & & \\text{Evaluate the integral}.\\hfill \\\\ & =\\frac{1}{9}{u}^{9}+\\frac{1}{7}{u}^{7}+C\\hfill & & \\text{Substitute }\\tan{x}=u.\\hfill \\\\ & =\\frac{1}{9}{\\tan}^{9}x+\\frac{1}{7}{\\tan}^{7}x+C.\\hfill & & \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      \r\n
      \r\n

      Evaluate [latex]{\\displaystyle\\int}{\\tan}^{5}x{\\sec}^{3}xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558099\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558099\"]\r\n

      \r\n

      Since the power on [latex]\\tan{x}[\/latex] is odd, begin by rewriting [latex]{\\tan}^{5}x{\\sec}^{3}x={\\tan}^{4}x{\\sec}^{2}x\\sec{x}\\tan{x}[\/latex]. Thus,<\/p>\r\n\r\n

      [latex]\\begin{array}{cccccc}\\hfill {\\tan}^{5}x{\\sec}^{3}x& =\\hfill & {\\tan}^{4}x{\\sec}^{2}x\\sec{x}\\tan{x}.\\hfill & & & \\text{Write }{\\tan}^{4}x={\\left({\\tan}^{2}x\\right)}^{2}.\\hfill \\\\ \\hfill {\\displaystyle\\int}{\\tan}^{5}x{\\sec}^{3}xdx& =\\hfill & {\\displaystyle\\int}{\\left({\\tan}^{2}x\\right)}^{2}{\\sec}^{2}x\\sec{x}\\tan{x}dx\\hfill & & & \\text{Use }{\\tan}^{2}x={\\sec}^{2}x - 1.\\hfill \\\\ & =\\hfill & {\\displaystyle\\int}{\\left({\\sec}^{2}x - 1\\right)}^{2}{\\sec}^{2}x\\sec{x}\\tan{x}dx\\hfill & & & \\text{Let }u=\\sec{x}\\text{and}du=\\sec{x}\\tan{x}dx.\\hfill \\\\ & =\\hfill & {\\displaystyle\\int}{\\left({u}^{2}-1\\right)}^{2}{u}^{2}du\\hfill & & & \\text{Expand}.\\hfill \\\\ & =\\hfill & {\\displaystyle\\int}\\left({u}^{6}-2{u}^{4}+{u}^{2}\\right)du\\hfill & & & \\text{Integrate}.\\hfill \\\\ & =\\hfill & \\frac{1}{7}{u}^{7}-\\frac{2}{5}{u}^{5}+\\frac{1}{3}{u}^{3}+C\\hfill & & & \\text{Substitute }\\sec{x}=u.\\hfill \\\\ & =\\hfill & \\frac{1}{7}{\\sec}^{7}x-\\frac{2}{5}{\\sec}^{5}x+\\frac{1}{3}{\\sec}^{3}x+C.\\hfill & & & \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      \r\n
      \r\n

      Integrate [latex]{\\displaystyle\\int}{\\sec}^{3}xdx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44556899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44556899\"]\r\n

      \r\n

      This integral requires integration by parts. To begin, let [latex]u=\\sec{x}[\/latex] and [latex]dv={\\sec}^{2}x[\/latex]. These choices make [latex]du=\\sec{x}\\tan{x}[\/latex] and [latex]v=\\tan{x}[\/latex]. Thus,<\/p>\r\n\r\n

      [latex]\\begin{array}{cccc}\\hfill {\\displaystyle\\int}{\\sec}^{3}xdx& =\\sec{x}\\tan{x}-{\\displaystyle\\int}\\tan{x}\\sec{x}\\tan{x}dx\\hfill & & \\\\ & =\\sec{x}\\tan{x}-{\\displaystyle\\int}{\\tan}^{2}x\\sec{x}dx\\hfill & & \\text{Simplify}.\\hfill \\\\ & =\\sec{x}\\tan{x}-{\\displaystyle\\int}\\left({\\sec}^{2}x - 1\\right)\\sec{x}dx\\hfill & & \\text{Substitute }{\\tan}^{2}x={\\sec}^{2}x - 1.\\hfill \\\\ & =\\sec{x}\\tan{x}+{\\displaystyle\\int}\\sec{x}dx-{\\displaystyle\\int}{\\sec}^{3}xdx\\hfill & & \\text{Rewrite}.\\hfill \\\\ & =\\sec{x}\\tan{x}+\\text{ln}|\\sec{x}+\\tan{x}|-{\\displaystyle\\int}{\\sec}^{3}xdx.\\hfill & & \\text{Evaluate}{\\displaystyle\\int}\\sec{x}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

      We now have<\/p>\r\n\r\n

      [latex]{\\displaystyle\\int}{\\sec}^{3}xdx=\\sec{x}\\tan{x}+\\text{ln}|\\sec{x}+\\tan{x}|-{\\displaystyle\\int}{\\sec}^{3}xdx[\/latex].<\/div>\r\n \r\n

      Since the integral [latex]{\\displaystyle\\int}{\\sec}^{3}xdx[\/latex] has reappeared on the right-hand side, we can solve for [latex]{\\displaystyle\\int}{\\sec}^{3}xdx[\/latex] by adding it to both sides. In doing so, we obtain<\/p>\r\n\r\n

      [latex]2{\\displaystyle\\int}{\\sec}^{3}xdx=\\sec{x}\\tan{x}+\\text{ln}|\\sec{x}+\\tan{x}|[\/latex].<\/div>\r\n \r\n

      Dividing by 2, we arrive at<\/p>\r\n\r\n

      [latex]{\\displaystyle\\int}{\\sec}^{3}xdx=\\frac{1}{2}\\sec{x}\\tan{x}+\\frac{1}{2}\\text{ln}|\\sec{x}+\\tan{x}|+C[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      [ohm_question hide_question_numbers=1]311294[\/ohm_question]<\/section>","rendered":"

      Integrating Products and Powers of [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex]<\/h2>\n

      Before discussing the integration of products and powers of [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex], it is useful to recall the integrals involving [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex] we have already learned:<\/p>\n

      \n
        \n
      1. [latex]{\\displaystyle\\int}{\\sec}^{2}xdx=\\tan{x}+C[\/latex]<\/li>\n
      2. [latex]{\\displaystyle\\int}\\sec{x}\\tan{x}dx=\\sec{x}+C[\/latex]<\/li>\n
      3. [latex]{\\displaystyle\\int}\\tan{x}dx=\\text{ln}|\\sec{x}|+C[\/latex]<\/li>\n
      4. [latex]{\\displaystyle\\int}\\sec{x}dx=\\text{ln}|\\sec{x}+\\tan{x}|+C[\/latex].<\/li>\n<\/ol>\n<\/section>\n

        For most integrals of products and powers of [latex]\\tan{x}[\/latex] and [latex]\\sec{x}[\/latex], we rewrite the expression we wish to integrate as the sum or difference of integrals of the form [latex]{\\displaystyle\\int}{\\tan}^{j}x{\\sec}^{2}xdx[\/latex] or [latex]{\\displaystyle\\int}{\\sec}^{j}x\\tan{x}dx[\/latex]. As we see in the following example, we can evaluate these new integrals by using u<\/em>-substitution.\u00a0 Before doing so, it is useful to note how the Pythagorean Identity implies relationships between other pairs of trigonometric functions.<\/p>\n

        \n

        For any angle [latex]x[\/latex]:<\/p>\n

        [latex]\\sin^2 x + \\cos^2 x = 1[\/latex]<\/div>\n

        Dividing the original equation by [latex]\\cos^2 x[\/latex] and simplifying yields an expression for [latex]\\sec^2 x[\/latex] in terms of [latex]\\tan^2 x[\/latex]:<\/p>\n

        [latex]\\tan^2 x + 1 = \\sec^2 x[\/latex]<\/div>\n

        Subtracting both sides of the equation by [latex]1[\/latex] yields an expression for [latex]\\tan^2 x[\/latex] in terms of [latex]\\sec^2 x[\/latex]:<\/p>\n

        [latex]\\tan^2 x = \\sec^2 x - 1[\/latex]<\/div>\n<\/section>\n
        \n
        \n

        Evaluate [latex]{\\displaystyle\\int}{\\sec}^{5}x\\tan{x}dx[\/latex].<\/p>\n<\/div>\n