{"id":1158,"date":"2021-06-30T17:02:02","date_gmt":"2021-06-30T17:02:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-integrals-resulting-in-inverse-trigonometric\/"},"modified":"2021-11-17T01:51:35","modified_gmt":"2021-11-17T01:51:35","slug":"problem-set-integrals-resulting-in-inverse-trigonometric","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/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

\r\n
\r\n

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

\r\n
\r\n

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

\r\n
\r\n

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

\r\n
\r\n

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

\r\n
\r\n

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

\r\n
\r\n

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

\n
\n

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
\n

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

\n
\n

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

\n
\n

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

Show Solution<\/span><\/p>\n
\n

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

\n
\n

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

\n
\n

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

Show Solution<\/span><\/p>\n
\n

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

\n
\n

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

\n\t\t\t

Candela Citations<\/h3>\n\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t\t
CC licensed content, Shared previously<\/div>