{"id":509,"date":"2021-02-04T15:37:57","date_gmt":"2021-02-04T15:37:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&p=509"},"modified":"2021-03-24T21:27:50","modified_gmt":"2021-03-24T21:27:50","slug":"problem-set-integrals-resulting-in-inverse-trigonometric","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/problem-set-integrals-resulting-in-inverse-trigonometric\/","title":{"raw":"Problem Set: Integrals Resulting in Inverse Trigonometric","rendered":"Problem Set: Integrals Resulting in Inverse Trigonometric"},"content":{"raw":"

In the following exercises (1-6), evaluate each integral in terms of an inverse trigonometric function.<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\frac{dx}{\\sqrt{1-{x}^{2}}}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572331724\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572331724\"]\r\n

[latex]{ \\sin }^{-1}x{|}_{0}^{\\sqrt{3}\\text{\/}2}=\\frac{\\pi }{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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2.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{-1\\text{\/}2}^{1\\text{\/}2}\\frac{dx}{\\sqrt{1-{x}^{2}}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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3.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{\\sqrt{3}}^{1}\\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572331752\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572331752\"]\r\n

[latex]{ \\tan }^{-1}x{|}_{\\sqrt{3}}^{1}=-\\frac{\\pi }{12}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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4.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{1\\text{\/}\\sqrt{3}}^{\\sqrt{3}}\\frac{dx}{1+{x}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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5.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{1}^{\\sqrt{2}}\\frac{dx}{|x|\\sqrt{{x}^{2}-1}}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572480541\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572480541\"]\r\n

[latex]{ \\sec }^{-1}x{|}_{1}^{\\sqrt{2}}=\\frac{\\pi }{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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6.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{1}^{2\\text{\/}\\sqrt{3}}\\frac{dx}{|x|\\sqrt{{x}^{2}-1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>","rendered":"

In the following exercises (1-6), evaluate each integral in terms of an inverse trigonometric function.<\/p>\n

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1.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{0}^{\\sqrt{3}\\text{\/}2}\\frac{dx}{\\sqrt{1-{x}^{2}}}[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]{ \\sin }^{-1}x{|}_{0}^{\\sqrt{3}\\text{\/}2}=\\frac{\\pi }{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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2.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{-1\\text{\/}2}^{1\\text{\/}2}\\frac{dx}{\\sqrt{1-{x}^{2}}}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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3.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{\\sqrt{3}}^{1}\\frac{dx}{\\sqrt{1+{x}^{2}}}[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]{ \\tan }^{-1}x{|}_{\\sqrt{3}}^{1}=-\\frac{\\pi }{12}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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4.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{1\\text{\/}\\sqrt{3}}^{\\sqrt{3}}\\frac{dx}{1+{x}^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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5.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{1}^{\\sqrt{2}}\\frac{dx}{|x|\\sqrt{{x}^{2}-1}}[\/latex]<\/p>\n

Show Solution<\/span><\/p>\n
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[latex]{ \\sec }^{-1}x{|}_{1}^{\\sqrt{2}}=\\frac{\\pi }{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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6.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{1}^{2\\text{\/}\\sqrt{3}}\\frac{dx}{|x|\\sqrt{{x}^{2}-1}}[\/latex]<\/p>\n<\/div>\n<\/div>\n\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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