{"id":1216,"date":"2025-07-16T15:17:06","date_gmt":"2025-07-16T15:17:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1216"},"modified":"2025-07-17T14:49:01","modified_gmt":"2025-07-17T14:49:01","slug":"trigonometric-integrals-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/trigonometric-integrals-learn-it-4\/","title":{"raw":"Trigonometric Integrals: Learn It 4","rendered":"Trigonometric Integrals: Learn It 4"},"content":{"raw":"
You've learned many integration techniques, but some expressions still seem impossible to tackle. Take [latex]\\int \\sqrt{9-{x}^{2}}dx[\/latex], for example. None of your usual methods work here\u2014but don't worry! There's a clever approach using trigonometric substitution.<\/p>\r\n
The trick is recognizing that expressions like [latex]\\sqrt{9-{x}^{2}}[\/latex] can be simplified using trigonometric identities. Let's see how this works with our example.<\/p>\r\n\r\n\r\n
Let's see how the substitution [latex]x=3\\sin\\theta[\/latex] transforms our integral [latex]\\displaystyle\\int \\sqrt{9-x^{2}}dx[\/latex].<\/p>\r\nIf we make the substitution [latex]x=3\\sin\\theta [\/latex], we have [latex]dx=3\\cos\\theta d\\theta [\/latex].\r\n\r\nAfter substituting into the integral, we have\r\n
At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions.<\/p>\r\n\r\n<\/section>Here's the systematic approach for any expression involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]:\r\n\r\n\r\n
the general method for [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\r\nThe Substitution:<\/strong> When you substitute [latex]x = a\\sin\\theta[\/latex]:\r\n\r\n
The domain of [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] is [latex][-a,a][\/latex]<\/li>\r\n \t
This means [latex]-1 \\leq \\frac{x}{a} \\leq 1[\/latex]<\/li>\r\n \t
Since [latex]\\sin\\theta[\/latex] has range [latex][-1,1][\/latex] over [latex][-\\frac{\\pi}{2}, \\frac{\\pi}{2}][\/latex], we can always find a [latex]\\theta[\/latex] where [latex]\\sin\\theta = \\frac{x}{a}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>After integrating, you need to express your answer in terms of [latex]x[\/latex]. This is where you'll need to recall some basic trigonometry to build a reference triangle. You'll use these relationships to convert your trigonometric answer back to an expression with [latex]x[\/latex].\r\n\r\n\r\n
Right Triangle Trigonometry<\/strong> For a right triangle with acute angle [latex]\\theta[\/latex]:<\/p>\r\n\r\n
Knowing [latex]x = a\\sin\\theta[\/latex], you can build a right triangle where [latex]\\sin\\theta = \\frac{x}{a}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] Figure 1. A reference triangle can help express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].[\/caption]\r\n
Since [latex]\\sin\\theta = \\frac{x}{a}[\/latex], we know:<\/p>\r\n\r\n
Check first:<\/strong> Can this integral be solved more easily another way? For example, although this method can be applied to integrals of the form [latex]\\displaystyle\\int \\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], [latex]\\displaystyle\\int \\frac{x}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], and [latex]\\displaystyle\\int x\\sqrt{{a}^{2}-{x}^{2}}dx[\/latex], they can each be integrated directly either by formula or by a simple u<\/em>-substitution.<\/li>\r\n \t
Substitute:<\/strong> Make the substitution [latex]x=a\\sin\\theta [\/latex] and [latex]dx=a\\cos\\theta d\\theta [\/latex].<\/li>\r\n \t
Simplify:<\/strong> Use [latex]\\sqrt{{a}^{2}-{x}^{2}} = a\\cos\\theta[\/latex]<\/li>\r\n \t
Integrate:<\/strong> Use trigonometric integration techniques<\/li>\r\n \t
Convert back<\/strong>: Use the reference triangle to express the result in terms of [latex]x[\/latex] You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\sin}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>This technique is most useful when you have [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] that can't be handled by simpler methods. However, some integrals like [latex]\\int \\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex] or [latex]\\int \\frac{x}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex] are better solved with basic formulas or u-substitution.<\/section>\r\n
Begin by making the substitutions [latex]x=3\\sin\\theta [\/latex] and [latex]dx=3\\cos\\theta d\\theta [\/latex]. Since [latex]\\sin\\theta =\\frac{x}{3}[\/latex], we can construct the reference triangle shown in the following figure.<\/p>\r\n\r\n<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] Figure 2. A reference triangle can be constructed for this example.[\/caption]<\/figure>\r\n
Thus,<\/p>\r\n\r\n
[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{x}^{2}}dx& ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9-{\\left(3\\sin\\theta \\right)}^{2}}3\\cos\\theta d\\theta \\hfill & & & \\text{Substitute }x=3\\sin\\theta \\text{ and }dx=3\\cos\\theta d\\theta .\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9\\left(1-{\\sin}^{2}\\theta \\right)}3\\cos\\theta d\\theta \\hfill & & & \\text{Simplify.}\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}\\sqrt{9{\\cos}^{2}\\theta }3\\cos\\theta d\\theta \\hfill & & & \\text{Substitute }{\\cos}^{2}\\theta =1-{\\sin}^{2}\\theta .\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}3|\\cos\\theta |3\\cos\\theta d\\theta \\hfill & & & \\text{Take the square root.}\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}9{\\cos}^{2}\\theta d\\theta \\hfill & & & \\begin{array}{c}\\text{Simplify. Since}-\\frac{\\pi }{2}\\le \\theta \\le \\frac{\\pi }{2},\\cos\\theta \\ge 0\\text{ and }\\hfill \\\\ |\\cos\\theta |=\\cos\\theta .\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int }^{\\text{ }}9\\left(\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2\\theta \\right)\\right)d\\theta \\hfill & & & \\begin{array}{c}\\text{Use the strategy for integrating an even power}\\hfill \\\\ \\text{of }\\cos\\theta .\\hfill \\end{array}\\hfill \\\\ & =\\frac{9}{2}\\theta +\\frac{9}{4}\\sin\\left(2\\theta \\right)+C\\hfill & & & \\text{Evaluate the integral.}\\hfill \\\\ & =\\frac{9}{2}\\theta +\\frac{9}{4}\\left(2\\sin\\theta \\cos\\theta \\right)+C\\hfill & & & \\text{Substitute }\\sin\\left(2\\theta \\right)=2\\sin\\theta \\cos\\theta .\\hfill \\\\ & =\\frac{9}{2}{\\sin}^{-1}\\left(\\frac{x}{3}\\right)+\\frac{9}{2}\\cdot \\frac{x}{3}\\cdot \\frac{\\sqrt{9-{x}^{2}}}{3}+C\\hfill & & & \\begin{array}{c}\\text{Substitute }{\\sin}^{-1}\\left(\\frac{x}{3}\\right)=\\theta \\text{ and }\\sin\\theta =\\frac{x}{3}.\\text{Use}\\hfill \\\\ \\text{the reference triangle to see that}\\hfill \\\\ \\cos\\theta =\\frac{\\sqrt{9-{x}^{2}}}{3}\\text{and make this substitution.}\\hfill \\end{array}\\hfill \\\\ & =\\frac{9}{2}{\\sin}^{-1}\\left(\\frac{x}{3}\\right)+\\frac{x\\sqrt{9-{x}^{2}}}{2}+C.\\hfill & & & \\text{Simplify.}\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>In the next example, we see that we sometimes have a choice of methods.\r\n\r\n\r\n
\r\n
Evaluate [latex]{\\displaystyle\\int }^{\\text{ }}{x}^{3}\\sqrt{1-{x}^{2}}dx[\/latex] two ways: first by using the substitution [latex]u=1-{x}^{2}[\/latex] and then by using a trigonometric substitution.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n
\r\n
Method 1<\/strong><\/p>\r\n
Let [latex]u=1-{x}^{2}[\/latex] and hence [latex]{x}^{2}=1-u[\/latex]. Thus, [latex]du=-2xdx[\/latex]. In this case, the integral becomes<\/p>\r\n\r\n
You’ve learned many integration techniques, but some expressions still seem impossible to tackle. Take [latex]\\int \\sqrt{9-{x}^{2}}dx[\/latex], for example. None of your usual methods work here\u2014but don’t worry! There’s a clever approach using trigonometric substitution.<\/p>\n
The trick is recognizing that expressions like [latex]\\sqrt{9-{x}^{2}}[\/latex] can be simplified using trigonometric identities. Let’s see how this works with our example.<\/p>\n\n
Let’s see how the substitution [latex]x=3\\sin\\theta[\/latex] transforms our integral [latex]\\displaystyle\\int \\sqrt{9-x^{2}}dx[\/latex].<\/p>\n
If we make the substitution [latex]x=3\\sin\\theta[\/latex], we have [latex]dx=3\\cos\\theta d\\theta[\/latex].<\/p>\n
After substituting into the integral, we have<\/p>\n
At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions.<\/p>\n<\/section>\n
Here’s the systematic approach for any expression involving [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]:<\/p>\n\n
the general method for [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex]<\/h3>\n
The Substitution:<\/strong> When you substitute [latex]x = a\\sin\\theta[\/latex]:<\/p>\n
The domain of [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] is [latex][-a,a][\/latex]<\/li>\n
This means [latex]-1 \\leq \\frac{x}{a} \\leq 1[\/latex]<\/li>\n
Since [latex]\\sin\\theta[\/latex] has range [latex][-1,1][\/latex] over [latex][-\\frac{\\pi}{2}, \\frac{\\pi}{2}][\/latex], we can always find a [latex]\\theta[\/latex] where [latex]\\sin\\theta = \\frac{x}{a}[\/latex]<\/li>\n<\/ul>\n<\/section>\n
After integrating, you need to express your answer in terms of [latex]x[\/latex]. This is where you’ll need to recall some basic trigonometry to build a reference triangle. You’ll use these relationships to convert your trigonometric answer back to an expression with [latex]x[\/latex].<\/p>\n\n
Right Triangle Trigonometry<\/strong> For a right triangle with acute angle [latex]\\theta[\/latex]:<\/p>\n
Knowing [latex]x = a\\sin\\theta[\/latex], you can build a right triangle where [latex]\\sin\\theta = \\frac{x}{a}[\/latex].<\/p>\nFigure 1. A reference triangle can help express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].<\/figcaption><\/figure>\n
Since [latex]\\sin\\theta = \\frac{x}{a}[\/latex], we know:<\/p>\n
Check first:<\/strong> Can this integral be solved more easily another way? For example, although this method can be applied to integrals of the form [latex]\\displaystyle\\int \\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], [latex]\\displaystyle\\int \\frac{x}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex], and [latex]\\displaystyle\\int x\\sqrt{{a}^{2}-{x}^{2}}dx[\/latex], they can each be integrated directly either by formula or by a simple u<\/em>-substitution.<\/li>\n
Substitute:<\/strong> Make the substitution [latex]x=a\\sin\\theta[\/latex] and [latex]dx=a\\cos\\theta d\\theta[\/latex].<\/li>\n
Simplify:<\/strong> Use [latex]\\sqrt{{a}^{2}-{x}^{2}} = a\\cos\\theta[\/latex]<\/li>\n
Integrate:<\/strong> Use trigonometric integration techniques<\/li>\n
Convert back<\/strong>: Use the reference triangle to express the result in terms of [latex]x[\/latex] You may also need to use some trigonometric identities and the relationship [latex]\\theta ={\\sin}^{-1}\\left(\\frac{x}{a}\\right)[\/latex].<\/li>\n<\/ol>\n<\/section>\nThis technique is most useful when you have [latex]\\sqrt{{a}^{2}-{x}^{2}}[\/latex] that can’t be handled by simpler methods. However, some integrals like [latex]\\int \\frac{1}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex] or [latex]\\int \\frac{x}{\\sqrt{{a}^{2}-{x}^{2}}}dx[\/latex] are better solved with basic formulas or u-substitution.<\/section>\n\n
![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233757\/CNX_Calc_Figure_07_03_001.jpg\")
Figure 1. A reference triangle can help express the trigonometric functions evaluated at [latex]\\theta [\/latex] in terms of [latex]x[\/latex].[\/caption]\r\n![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233800\/CNX_Calc_Figure_07_03_002.jpg\")
Figure 2. A reference triangle can be constructed for this example.[\/caption]<\/figure>\r\n