{"id":157,"date":"2026-01-12T15:55:57","date_gmt":"2026-01-12T15:55:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&p=157"},"modified":"2026-01-12T15:55:58","modified_gmt":"2026-01-12T15:55:58","slug":"techniques-for-integration-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/techniques-for-integration-get-stronger-answer-key\/","title":{"raw":"Techniques for Integration: Get Stronger Answer Key","rendered":"Techniques for Integration: Get Stronger Answer Key"},"content":{"raw":"

Substitution<\/h2>\r\n
    \r\n \t
  1. <\/li>\r\n \t
  2. [latex]u=h(x)[\/latex]<\/li>\r\n<li[latex]f(u)=\\frac{{(u+1)}^{2}}{\\sqrt{u}}[\/latex]\r\n \t
  3. [latex]du=8xdx;f(u)=\\frac{1}{8\\sqrt{u}}[\/latex]<\/li>\r\n \t
  4. [latex]\\frac{1}{5}{(x+1)}^{5}+C[\/latex]<\/li>\r\n \t
  5. [latex]-\\frac{1}{12{(3-2x)}^{6}}+C[\/latex]<\/li>\r\n \t
  6. [latex]\\sqrt{{x}^{2}+1}+C[\/latex]<\/li>\r\n \t
  7. [latex]\\frac{1}{8}{({x}^{2}-2x)}^{4}+C[\/latex]<\/li>\r\n \t
  8. [latex] \\sin \\theta -\\frac{{ \\sin }^{3}\\theta }{3}+C[\/latex]<\/li>\r\n \t
  9. [latex]\\frac{{(1-x)}^{101}}{101}-\\frac{{(1-x)}^{100}}{100}+C[\/latex]<\/li>\r\n \t
  10. [latex]-\\frac{1}{22(7-11{x}^{2})}+C[\/latex]<\/li>\r\n \t
  11. [latex]-\\frac{{ \\cos }^{4}\\theta }{4}+C[\/latex]<\/li>\r\n \t
  12. [latex]-\\frac{{ \\cos }^{3}(\\pi t)}{3\\pi }+C[\/latex]<\/li>\r\n \t
  13. [latex]-\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\cos }^{2}({t}^{2})+C[\/latex]<\/li>\r\n \t
  14. [latex]-\\frac{1}{3({x}^{3}-3)}+C[\/latex]<\/li>\r\n \t
  15. [latex]-\\frac{2({y}^{3}-2)}{3\\sqrt{1-{y}^{3}}}[\/latex]<\/li>\r\n \t
  16. [latex]\\frac{1}{33}{(1-{ \\cos }^{3}\\theta )}^{11}+C[\/latex]<\/li>\r\n \t
  17. [latex]\\frac{1}{12}{({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{4}+C[\/latex]<\/li>\r\n \t
  18. [latex]{L}_{50}=-8.5779.[\/latex] The exact area is [latex]\\frac{-81}{8}[\/latex]<\/li>\r\n \t
  19. [latex]{L}_{50}=-0.006399[\/latex] \u2026 The exact area is [latex]0[\/latex].<\/li>\r\n \t
  20. [latex]u=1+{x}^{2},du=2xdx,\\frac{1}{2}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\sqrt{2}-1[\/latex]<\/li>\r\n \t
  21. [latex]u=1+{t}^{3},du=3{t}^{2}du,\\frac{1}{3}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\frac{2}{3}(\\sqrt{2}-1)[\/latex]<\/li>\r\n \t
  22. [latex]u= \\cos \\theta ,du=\\text{\u2212} \\sin \\theta d\\theta ,{\\displaystyle\\int }_{1\\text{\/}\\sqrt{2}}^{1}{u}^{-4}du=\\frac{1}{3}(2\\sqrt{2}-1)[\/latex]<\/li>\r\n \t
  23. \"Two\r\nThe antiderivative is [latex]y= \\sin (\\text{ln}(2x)).[\/latex] Since the antiderivative is not continuous at [latex]x=0,[\/latex] one cannot find a value of C<\/em> that would make [latex]y= \\sin (\\text{ln}(2x))-C[\/latex] work as a definite integral.<\/li>\r\n \t
  24. \"Two\r\nThe antiderivative is [latex]y=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{2}x.[\/latex] You should take [latex]C=-2[\/latex] so that [latex]F(-\\frac{\\pi }{3})=0.[\/latex]<\/li>\r\n \t
  25. \"Two\r\nThe antiderivative is [latex]y=\\frac{1}{3}{(2{x}^{3}+1)}^{3\\text{\/}2}.[\/latex] One should take [latex]C=-\\frac{1}{3}.[\/latex]<\/li>\r\n \t
  26. No, because the integrand is discontinuous at [latex]x=1.[\/latex]<\/li>\r\n \t
  27. [latex]u= \\sin ({t}^{2});[\/latex] the integral becomes [latex]\\frac{1}{2}{\\displaystyle\\int }_{0}^{0}udu.[\/latex]<\/li>\r\n \t
  28. [latex]u=(1+{(t-\\frac{1}{2})}^{2});[\/latex] the integral becomes [latex]\\text{\u2212}{\\displaystyle\\int }_{5\\text{\/}4}^{5\\text{\/}4}\\frac{1}{u}du.[\/latex]<\/li>\r\n \t
  29. [latex]u=1-t;[\/latex] the integral becomes[latex]\\begin{array}{l}{\\displaystyle\\int }_{1}^{-1}u \\cos (\\pi (1-u))du\\hfill \\\\ ={\\displaystyle\\int }_{1}^{-1}u\\left[ \\cos \\pi \\cos u- \\sin \\pi \\sin u\\right]du\\hfill \\\\ =\\text{\u2212}{\\displaystyle\\int }_{1}^{-1}u \\cos udu\\hfill \\\\ ={\\displaystyle\\int }_{-1}^{1}u \\cos udu=0\\hfill \\end{array}[\/latex]\r\nsince the integrand is odd.<\/li>\r\n \t
  30. Setting [latex]u=cx[\/latex] and [latex]du=cdx[\/latex] gets you [latex]\\frac{1}{\\frac{b}{c}-\\frac{a}{c}}{\\displaystyle\\int }_{a\\text{\/}c}^{b\\text{\/}c}f(cx)dx=\\frac{c}{b-a}{\\displaystyle\\int }_{u=a}^{u=b}f(u)\\frac{du}{c}=\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(u)du.[\/latex]<\/li>\r\n \t
  31. [latex]{\\displaystyle\\int }_{0}^{x}g(t)dt=\\frac{1}{2}{\\displaystyle\\int }_{u=1-{x}^{2}}^{1}\\frac{du}{{u}^{a}}=\\frac{1}{2(1-a)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\\frac{1}{2(1-a)}(1-{(1-{x}^{2})}^{1-a}).[\/latex] As [latex]x\\to 1[\/latex] the limit is [latex]\\frac{1}{2(1-a)}[\/latex] if [latex]a<1,[\/latex] and the limit diverges to +\u221e if [latex]a>1.[\/latex]<\/li>\r\n \t
  32. [latex]{\\displaystyle\\int }_{t=\\pi }^{0}b\\sqrt{1-{ \\cos }^{2}t}\u00d7(\\text{\u2212}a \\sin t)dt={\\displaystyle\\int }_{t=0}^{\\pi }ab{ \\sin }^{2}tdt[\/latex]<\/li>\r\n \t
  33. [latex]f(t)=2 \\cos (3t)- \\cos (2t);{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}(2 \\cos (3t)- \\cos (2t))=-\\frac{2}{3}[\/latex]<\/li>\r\n<\/ol>\r\n

    Integrals Involving Exponential and Logarithmic Functions<\/h2>\r\n
      \r\n \t
    1. [latex]\\frac{1}{9}{x}^{3}(\\text{ln}({x}^{3})-1)+C[\/latex]<\/li>\r\n \t
    2. [latex]2\\sqrt{x}(\\text{ln}x-2)+C[\/latex]<\/li>\r\n \t
    3. [latex]{\\displaystyle\\int }_{0}^{\\text{ln}x}{e}^{t}dt={e}^{t}{|}_{0}^{\\text{ln}x}={e}^{\\text{ln}x}-{e}^{0}=x-1[\/latex]<\/li>\r\n \t
    4. [latex]-\\frac{1}{3}\\text{ln}( \\sin (3x)+ \\cos (3x))[\/latex]<\/li>\r\n \t
    5. [latex]-\\frac{1}{2}\\text{ln}| \\csc ({x}^{2})+ \\cot ({x}^{2})|+C[\/latex]<\/li>\r\n \t
    6. [latex]-\\frac{1}{2}{(\\text{ln}( \\csc x))}^{2}+C[\/latex]<\/li>\r\n \t
    7. [latex]\\frac{1}{3}\\text{ln}(\\frac{26}{7})[\/latex]<\/li>\r\n \t
    8. [latex]y-2\\text{ln}|y+1|+C[\/latex]<\/li>\r\n \t
    9. [latex]\\text{ln}| \\sin x- \\cos x|+C[\/latex]<\/li>\r\n \t
    10. [latex]-\\frac{1}{3}{(1-(\\text{ln}{x}^{2}))}^{3\\text{\/}2}+C[\/latex]<\/li>\r\n \t
    11. Exact solution: [latex]\\frac{e-1}{e},{R}_{50}=0.6258.[\/latex] Since [latex]f[\/latex] is decreasing, the right endpoint estimate underestimates the area.<\/li>\r\n \t
    12. Exact solution: [latex]\\frac{2\\text{ln}(3)-\\text{ln}(6)}{2},{R}_{50}=0.2033.[\/latex] Since [latex]f[\/latex] is increasing, the right endpoint estimate overestimates the area.<\/li>\r\n \t
    13. Exact solution: [latex]-\\frac{1}{\\text{ln}(4)},{R}_{50}=-0.7164.[\/latex] Since [latex]f[\/latex] is increasing, the right endpoint estimate overestimates the area (the actual area is a larger negative number).<\/li>\r\n<\/ol>\r\n

      Integrals Resulting in Inverse Trigonometric<\/h2>\r\n
        \r\n \t
      1. [latex]{ \\sin }^{-1}x{|}_{0}^{\\sqrt{3}\\text{\/}2}=\\frac{\\pi }{3}[\/latex]<\/li>\r\n \t
      2. [latex]{ \\tan }^{-1}x{|}_{\\sqrt{3}}^{1}=-\\frac{\\pi }{12}[\/latex]<\/li>\r\n \t
      3. [latex]{ \\sec }^{-1}x{|}_{1}^{\\sqrt{2}}=\\frac{\\pi }{4}[\/latex]<\/li>\r\n<\/ol>","rendered":"

        Substitution<\/h2>\n
          \n
        1. <\/li>\n
        2. [latex]u=h(x)[\/latex]<\/li>\n
        3. [latex]du=8xdx;f(u)=\\frac{1}{8\\sqrt{u}}[\/latex]<\/li>\n
        4. [latex]\\frac{1}{5}{(x+1)}^{5}+C[\/latex]<\/li>\n
        5. [latex]-\\frac{1}{12{(3-2x)}^{6}}+C[\/latex]<\/li>\n
        6. [latex]\\sqrt{{x}^{2}+1}+C[\/latex]<\/li>\n
        7. [latex]\\frac{1}{8}{({x}^{2}-2x)}^{4}+C[\/latex]<\/li>\n
        8. [latex]\\sin \\theta -\\frac{{ \\sin }^{3}\\theta }{3}+C[\/latex]<\/li>\n
        9. [latex]\\frac{{(1-x)}^{101}}{101}-\\frac{{(1-x)}^{100}}{100}+C[\/latex]<\/li>\n
        10. [latex]-\\frac{1}{22(7-11{x}^{2})}+C[\/latex]<\/li>\n
        11. [latex]-\\frac{{ \\cos }^{4}\\theta }{4}+C[\/latex]<\/li>\n
        12. [latex]-\\frac{{ \\cos }^{3}(\\pi t)}{3\\pi }+C[\/latex]<\/li>\n
        13. [latex]-\\frac{1}{4}\\phantom{\\rule{0.05em}{0ex}}{ \\cos }^{2}({t}^{2})+C[\/latex]<\/li>\n
        14. [latex]-\\frac{1}{3({x}^{3}-3)}+C[\/latex]<\/li>\n
        15. [latex]-\\frac{2({y}^{3}-2)}{3\\sqrt{1-{y}^{3}}}[\/latex]<\/li>\n
        16. [latex]\\frac{1}{33}{(1-{ \\cos }^{3}\\theta )}^{11}+C[\/latex]<\/li>\n
        17. [latex]\\frac{1}{12}{({ \\sin }^{3}\\theta -3{ \\sin }^{2}\\theta )}^{4}+C[\/latex]<\/li>\n
        18. [latex]{L}_{50}=-8.5779.[\/latex] The exact area is [latex]\\frac{-81}{8}[\/latex]<\/li>\n
        19. [latex]{L}_{50}=-0.006399[\/latex] \u2026 The exact area is [latex]0[\/latex].<\/li>\n
        20. [latex]u=1+{x}^{2},du=2xdx,\\frac{1}{2}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\sqrt{2}-1[\/latex]<\/li>\n
        21. [latex]u=1+{t}^{3},du=3{t}^{2}du,\\frac{1}{3}{\\displaystyle\\int }_{1}^{2}{u}^{-1\\text{\/}2}du=\\frac{2}{3}(\\sqrt{2}-1)[\/latex]<\/li>\n
        22. [latex]u= \\cos \\theta ,du=\\text{\u2212} \\sin \\theta d\\theta ,{\\displaystyle\\int }_{1\\text{\/}\\sqrt{2}}^{1}{u}^{-4}du=\\frac{1}{3}(2\\sqrt{2}-1)[\/latex]<\/li>\n
        23. \"Two
          \nThe antiderivative is [latex]y= \\sin (\\text{ln}(2x)).[\/latex] Since the antiderivative is not continuous at [latex]x=0,[\/latex] one cannot find a value of C<\/em> that would make [latex]y= \\sin (\\text{ln}(2x))-C[\/latex] work as a definite integral.<\/li>\n
        24. \"Two
          \nThe antiderivative is [latex]y=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\sec }^{2}x.[\/latex] You should take [latex]C=-2[\/latex] so that [latex]F(-\\frac{\\pi }{3})=0.[\/latex]<\/li>\n
        25. \"Two
          \nThe antiderivative is [latex]y=\\frac{1}{3}{(2{x}^{3}+1)}^{3\\text{\/}2}.[\/latex] One should take [latex]C=-\\frac{1}{3}.[\/latex]<\/li>\n
        26. No, because the integrand is discontinuous at [latex]x=1.[\/latex]<\/li>\n
        27. [latex]u= \\sin ({t}^{2});[\/latex] the integral becomes [latex]\\frac{1}{2}{\\displaystyle\\int }_{0}^{0}udu.[\/latex]<\/li>\n
        28. [latex]u=(1+{(t-\\frac{1}{2})}^{2});[\/latex] the integral becomes [latex]\\text{\u2212}{\\displaystyle\\int }_{5\\text{\/}4}^{5\\text{\/}4}\\frac{1}{u}du.[\/latex]<\/li>\n
        29. [latex]u=1-t;[\/latex] the integral becomes[latex]\\begin{array}{l}{\\displaystyle\\int }_{1}^{-1}u \\cos (\\pi (1-u))du\\hfill \\\\ ={\\displaystyle\\int }_{1}^{-1}u\\left[ \\cos \\pi \\cos u- \\sin \\pi \\sin u\\right]du\\hfill \\\\ =\\text{\u2212}{\\displaystyle\\int }_{1}^{-1}u \\cos udu\\hfill \\\\ ={\\displaystyle\\int }_{-1}^{1}u \\cos udu=0\\hfill \\end{array}[\/latex]
          \nsince the integrand is odd.<\/li>\n
        30. Setting [latex]u=cx[\/latex] and [latex]du=cdx[\/latex] gets you [latex]\\frac{1}{\\frac{b}{c}-\\frac{a}{c}}{\\displaystyle\\int }_{a\\text{\/}c}^{b\\text{\/}c}f(cx)dx=\\frac{c}{b-a}{\\displaystyle\\int }_{u=a}^{u=b}f(u)\\frac{du}{c}=\\frac{1}{b-a}{\\displaystyle\\int }_{a}^{b}f(u)du.[\/latex]<\/li>\n
        31. [latex]{\\displaystyle\\int }_{0}^{x}g(t)dt=\\frac{1}{2}{\\displaystyle\\int }_{u=1-{x}^{2}}^{1}\\frac{du}{{u}^{a}}=\\frac{1}{2(1-a)}{u}^{1-a}{|}_{u=1-{x}^{2}}^{1}=\\frac{1}{2(1-a)}(1-{(1-{x}^{2})}^{1-a}).[\/latex] As [latex]x\\to 1[\/latex] the limit is [latex]\\frac{1}{2(1-a)}[\/latex] if [latex]a<1,[\/latex] and the limit diverges to +\u221e if [latex]a>1.[\/latex]<\/li>\n
        32. [latex]{\\displaystyle\\int }_{t=\\pi }^{0}b\\sqrt{1-{ \\cos }^{2}t}\u00d7(\\text{\u2212}a \\sin t)dt={\\displaystyle\\int }_{t=0}^{\\pi }ab{ \\sin }^{2}tdt[\/latex]<\/li>\n
        33. [latex]f(t)=2 \\cos (3t)- \\cos (2t);{\\displaystyle\\int }_{0}^{\\pi \\text{\/}2}(2 \\cos (3t)- \\cos (2t))=-\\frac{2}{3}[\/latex]<\/li>\n<\/ol>\n

          Integrals Involving Exponential and Logarithmic Functions<\/h2>\n
            \n
          1. [latex]\\frac{1}{9}{x}^{3}(\\text{ln}({x}^{3})-1)+C[\/latex]<\/li>\n
          2. [latex]2\\sqrt{x}(\\text{ln}x-2)+C[\/latex]<\/li>\n
          3. [latex]{\\displaystyle\\int }_{0}^{\\text{ln}x}{e}^{t}dt={e}^{t}{|}_{0}^{\\text{ln}x}={e}^{\\text{ln}x}-{e}^{0}=x-1[\/latex]<\/li>\n
          4. [latex]-\\frac{1}{3}\\text{ln}( \\sin (3x)+ \\cos (3x))[\/latex]<\/li>\n
          5. [latex]-\\frac{1}{2}\\text{ln}| \\csc ({x}^{2})+ \\cot ({x}^{2})|+C[\/latex]<\/li>\n
          6. [latex]-\\frac{1}{2}{(\\text{ln}( \\csc x))}^{2}+C[\/latex]<\/li>\n
          7. [latex]\\frac{1}{3}\\text{ln}(\\frac{26}{7})[\/latex]<\/li>\n
          8. [latex]y-2\\text{ln}|y+1|+C[\/latex]<\/li>\n
          9. [latex]\\text{ln}| \\sin x- \\cos x|+C[\/latex]<\/li>\n
          10. [latex]-\\frac{1}{3}{(1-(\\text{ln}{x}^{2}))}^{3\\text{\/}2}+C[\/latex]<\/li>\n
          11. Exact solution: [latex]\\frac{e-1}{e},{R}_{50}=0.6258.[\/latex] Since [latex]f[\/latex] is decreasing, the right endpoint estimate underestimates the area.<\/li>\n
          12. Exact solution: [latex]\\frac{2\\text{ln}(3)-\\text{ln}(6)}{2},{R}_{50}=0.2033.[\/latex] Since [latex]f[\/latex] is increasing, the right endpoint estimate overestimates the area.<\/li>\n
          13. Exact solution: [latex]-\\frac{1}{\\text{ln}(4)},{R}_{50}=-0.7164.[\/latex] Since [latex]f[\/latex] is increasing, the right endpoint estimate overestimates the area (the actual area is a larger negative number).<\/li>\n<\/ol>\n

            Integrals Resulting in Inverse Trigonometric<\/h2>\n
              \n
            1. [latex]{ \\sin }^{-1}x{|}_{0}^{\\sqrt{3}\\text{\/}2}=\\frac{\\pi }{3}[\/latex]<\/li>\n
            2. [latex]{ \\tan }^{-1}x{|}_{\\sqrt{3}}^{1}=-\\frac{\\pi }{12}[\/latex]<\/li>\n
            3. [latex]{ \\sec }^{-1}x{|}_{1}^{\\sqrt{2}}=\\frac{\\pi }{4}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":108,"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\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/157"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":1,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/157\/revisions"}],"predecessor-version":[{"id":170,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/157\/revisions\/170"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/108"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/157\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=157"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=157"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=157"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}