{"id":413,"date":"2025-02-13T19:44:43","date_gmt":"2025-02-13T19:44:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-resulting-in-inverse-trigonometric-functions-learn-it-2\/"},"modified":"2025-02-13T19:44:43","modified_gmt":"2025-02-13T19:44:43","slug":"integrals-resulting-in-inverse-trigonometric-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integrals-resulting-in-inverse-trigonometric-functions-learn-it-2\/","title":{"raw":"Integrals Resulting in Inverse Trigonometric Functions: Learn It 2","rendered":"Integrals Resulting in Inverse Trigonometric Functions: Learn It 2"},"content":{"raw":"\n

Integrals Resulting in Other Inverse Trigonometric Functions <\/h2>\n

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative.<\/p>\n

\n

Rather than memorizing three more formulas, if the integrand is negative, simply factor out [latex]\u22121[\/latex] and evaluate the integral using one of the formulas already provided.<\/p>\n<\/section>\n

\n

Find an antiderivative of [latex]\\displaystyle\\int \\frac{1}{1+4{x}^{2}}dx.[\/latex]<\/p>\n

[reveal-answer q=\"fs-id1170572449549\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1170572449549\"]\n\n

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for [latex]{ \\tan }^{-1}u+C.[\/latex]<\/p>\n

So we use substitution, letting [latex]u=2x,[\/latex] then [latex]du=2dx[\/latex] and [latex]\\frac{1}{2}du=dx.[\/latex]<\/p>\n

Then, we have,<\/p>\n

[latex]\\frac{1}{2}\\displaystyle\\int \\frac{1}{1+{u}^{2}}du=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}u+C=\\frac{1}{2}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}(2x)+C.[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/section>\n
\n

Find the antiderivative of [latex]\\displaystyle\\int \\frac{1}{9+{x}^{2}}dx.[\/latex]<\/p>\n

\n
\n

[reveal-answer q=\"7084662\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"7084662\"]<\/p>\n

Apply the formula with [latex]a=3.[\/latex] Then,<\/p>\n

[latex]\\displaystyle\\int \\frac{dx}{9+{x}^{2}}=\\frac{1}{3}\\phantom{\\rule{0.05em}{0ex}}{ \\tan }^{-1}\\left(\\frac{x}{3}\\right)+C[\/latex]<\/div>\n
[\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/section>\n
\n

Evaluate the definite integral [latex]{\\displaystyle\\int }_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\dfrac{dx}{1+{x}^{2}}.[\/latex]<\/p>\n

[reveal-answer q=\"fs-id1170572099768\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1170572099768\"]\n\n

Use the formula for the inverse tangent. We have,<\/p>\n

[latex]\\begin{array}{} \\\\ {\\displaystyle\\int }_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\dfrac{dx}{1+{x}^{2}}\\hfill & ={ \\tan }^{-1}x{|}_{\\sqrt{3}\\text{\/}3}^{\\sqrt{3}}\\hfill \\\\ & =\\left[{ \\tan }^{-1}(\\sqrt{3})\\right]-\\left[{ \\tan }^{-1}\\left(\\frac{\\sqrt{3}}{3}\\right)\\right]\\hfill \\\\ & =\\frac{\\pi }{6}.\\hfill \\end{array}[\/latex]<\/div>\n
 <\/div>\n
[\/hidden-answer] <\/div>\n<\/div>\n<\/section>\n
\n

[ohm_question hide_question_numbers=1]288438[\/ohm_question]<\/p>\n<\/section>\n","rendered":"

Integrals Resulting in Other Inverse Trigonometric Functions <\/h2>\n

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative.<\/p>\n

\n

Rather than memorizing three more formulas, if the integrand is negative, simply factor out [latex]\u22121[\/latex] and evaluate the integral using one of the formulas already provided.<\/p>\n<\/section>\n

\n

Find an antiderivative of [latex]\\displaystyle\\int \\frac{1}{1+4{x}^{2}}dx.[\/latex]<\/p>\n

\n