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17Calculus Integrals - Secant/Cosecant Substitution

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Secant/Cosecant Substitution

This is a great technique that introduces secant or cosecant into an integral which originally doesn't have any trig. This makes integration easier and sometimes possible where it wasn't before.
Make sure you have throughly gone through the trig substitution page since this page builds on that discussion.

Once you have determined that you have an integrand containing an \((x^2 - a^2)\) term, you use these equations for the substitution.

Secant Substitution

\(a\) is a constant; \(x\) and \(\theta\) are variables

integrand

\(x^2 - a^2\)

substitution

\(x = a~\sec(\theta)\)

identity

\( 1+\tan^2(\theta) = \sec^2(\theta)\)

Cosecant Substitution

Some books and instructors may use cosecant substitution instead of secant. In this case, the substitution is \(x = a\csc(\theta)\) and the identity is \( 1+\cot^2(\theta) = \csc^2(\theta) \).

In this case, we have two choices, to use secant or cosecant. Fortunately, both should work. So you may choose either based on what you are comfortable with, what your instructor requires and any other constraints. We will discuss the use of cosecant but the use of secant parallels this and we think you can easily extrapolate and alter the discussion to fit secant.

Similar to the other two cases above, we have the equation \(1+\cot^2(\theta) = \csc^2(\theta)\) that we will use to simplify. Performing the substitution, we have

\(\begin{array}{rcl} -a^2+x^2 & = & -a^2 + (a\csc(\theta))^2 \\ & = & -a^2 + a^2 \csc^2(\theta) \\ & = & a^2(-1+\csc^2(\theta)) \\ & = & a^2 \cot^2(\theta) \end{array}\)

This is a much simpler expression, especially if the \(-a^2+x^2\) is under a square root.

We use \(x = a~\csc(\theta) \to \csc(\theta) = x/a \to 1/\sin(\theta) = x/a \to \sin(\theta) = a/x \)   to build the triangle on the right. We choose one angle to label \(\theta\), label the other two sides using the last equation and then calculate the expression for the third side using the Pythagorean Theorem.

Next, we substitute in the integral for \(x = a\csc(\theta)\) and \(dx = a\csc(\theta)\cot(\theta)~d\theta \). This last substitution is important to remember since we can't just replace \(dx\) with \(d\theta\) and it is a common source of error for many students.

Finally, once we are done integrating, we use the same triangle above to convert all the terms back into \(x\). It is not correct to leave \(\theta\)'s in the final answer.

Okay, here are some practice problems.

Practice

Evaluate these integrals using secant or cosecant substitution. Give your answers in exact, simplified and factored form.

Basic

\(\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } = \ln \left| \frac{8}{5\sqrt{3}+\sqrt{59}} \right| }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

Krista King Math - 63 video solution

video by Krista King Math

Final Answer

\(\displaystyle{ \int_{10\sqrt{3}}^{10}{ \frac{1}{\sqrt{x^2-64}} ~dx } = \ln \left| \frac{8}{5\sqrt{3}+\sqrt{59}} \right| }\)

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\(\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } = \ln \left| x + \sqrt{x^2-4} \right| + C }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

PatrickJMT - 66 video solution

video by PatrickJMT

Final Answer

\(\displaystyle{ \int{ \frac{dx}{\sqrt{x^2-4}} } = \ln \left| x + \sqrt{x^2-4} \right| + C }\)

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Intermediate

\(\displaystyle{ \int{ \frac{1}{x^2\sqrt{x^2-9}} ~dx } }\)

Problem Statement

\(\displaystyle{ \int{ \frac{1}{x^2\sqrt{x^2-9}} ~dx } }\)

Solution

Michael Penn - 4141 video solution

video by Michael Penn

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\(\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } = \frac{\sqrt{3}}{32} }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

This problem is solved in three consecutive videos.

Krista King Math - 73 video solution

video by Krista King Math

Krista King Math - 73 video solution

video by Krista King Math

Krista King Math - 73 video solution

video by Krista King Math

Final Answer

\(\displaystyle{ \int_{4}^{8}{ \frac{dx}{x^2\sqrt{x^2-16}} } = \frac{\sqrt{3}}{32} }\)

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\(\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \sqrt{x^2-25}-5\text{arcsec}(x/5) + C }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{\sqrt{x^2-25}}{x} ~dx } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

Krista King Math - 74 video solution

video by Krista King Math

Final Answer

\(\displaystyle{ \sqrt{x^2-25}-5\text{arcsec}(x/5) + C }\)

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Expanding The Use of The Secant/Cosecant Substitution

In the discussion so far on this page, we have assumed that the \(x\) variable term is always exactly \(x\). There is a generalization of this where you might have \([f(x)]^2 - a^2\) and \(f(x) \neq x\). This technique still works and you just change the substitution as follows.

Expanded Secant/Cosecant Substitution

integrand

 

substitution

\([f(x)]^2-a^2\)

 

\(f(x) = a\sec(\theta)\)

\([f(x)]^2-a^2\)

 

\(f(x) = a\csc(\theta)\)

Example

\(4x^2-25\)

 

\(2x = 5\sec(\theta)\)

\(f(x)\) can be pretty much anything except for a constant. So check that it contains at least one variable. In the example above, \(f(x)=2x\).

Note: The integrand may not be given in the exact form \( [f(x)]^2 - a^2 \). You may need to complete the square in order to get this form.

Okay, let's work some practice problems using these ideas. You don't want to work these problems until you have mastered the basic technique in the previous section on this page.

Practice

Evaluate these integrals using secant or cosecant substitution. Give your answers in exact, simplified and factored form.

\(\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} }\)

Problem Statement

Evaluate \(\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \frac{1}{2}(x+2)\sqrt{x^2+4x+5} + }\) \(\displaystyle{ \frac{1}{2}\ln\left| \sqrt{x^2+4x+5}+x+2 \right| + C }\)

Problem Statement

Evaluate \(\displaystyle{ \int{\sqrt{x^2+4x+5}~dx} }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

Krista King Math - 78 video solution

video by Krista King Math

Final Answer

\(\displaystyle{ \frac{1}{2}(x+2)\sqrt{x^2+4x+5} + }\) \(\displaystyle{ \frac{1}{2}\ln\left| \sqrt{x^2+4x+5}+x+2 \right| + C }\)

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\(\displaystyle{ \int{ \frac{dx}{(4x^2-1)^{3/2}} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{(4x^2-1)^{3/2}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \int{ \frac{dx}{(4x^2-1)^{3/2}} } = \frac{-x}{\sqrt{4x^2-1}} + C }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{(4x^2-1)^{3/2}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

Krista King Math - 72 video solution

video by Krista King Math

Final Answer

\(\displaystyle{ \int{ \frac{dx}{(4x^2-1)^{3/2}} } = \frac{-x}{\sqrt{4x^2-1}} + C }\)

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\(\displaystyle{ \int{ \frac{dx}{x^2+4x+1} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{x^2+4x+1} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

PatrickJMT - 1851 video solution

video by PatrickJMT

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\(\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \frac{-\sqrt{15}}{30} \left[ \ln\left| x-4 + \sqrt{15} \right| - \ln \left| x-4-\sqrt{15} \right| ~ \right] + C }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{x^2-8x+1} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

The instructor does not complete the problem in the video but does lay out how to solve it. Here is the rest of the solution.

\(\displaystyle{ \frac{\sqrt{15}}{15}\int{\csc \theta ~d\theta} = \frac{-\sqrt{15}}{15}\ln\left| \csc\theta + \cot\theta \right| + C }\)
Note: This integral is solved in a practice problem on the trig integration page.
Now we need to convert back to x terms. Using the triangle from the video, we have

\(\displaystyle{ \csc\theta = \frac{1}{\sin\theta} = \frac{x-4}{\sqrt{(x-4)^2-15}} }\)

\(\displaystyle{ \cot \theta = \frac{\cos\theta}{\sin\theta} = \frac{\sqrt{15}}{\sqrt{(x-4)^2-15}} }\)

\(\displaystyle{ \frac{-\sqrt{15}}{15}\ln\left| \csc\theta + \cot\theta \right| + C }\)

\(\displaystyle{ \frac{-\sqrt{15}}{15}\ln\left| \frac{x-4}{\sqrt{(x-4)^2-15}} + \right. }\) \(\displaystyle{ \left. \frac{\sqrt{15}}{\sqrt{(x-4)^2-15}} \right| + C }\)

\(\displaystyle{ \frac{-\sqrt{15}}{15}\ln\left| \frac{x-4 + \sqrt{15}}{\sqrt{(x-4)^2-15}} \right| + C }\)

Note: This answer can be simplified more to

\(\displaystyle{ \frac{-\sqrt{15}}{30}\ln\left| \frac{ x-4 + \sqrt{15}}{x-4-\sqrt{15}} \right| + C = }\) \(\displaystyle{ \frac{-\sqrt{15}}{30} \left[ \ln\left| x-4 + \sqrt{15} \right| - \ln \left| x-4-\sqrt{15} \right| ~ \right] + C }\)

Check with your instructor to see how much simplification they require.

MIT OCW - 76 video solution

video by MIT OCW

Final Answer

\(\displaystyle{ \frac{-\sqrt{15}}{30} \left[ \ln\left| x-4 + \sqrt{15} \right| - \ln \left| x-4-\sqrt{15} \right| ~ \right] + C }\)

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\(\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Final Answer

\(\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } = \frac{-1}{25} \left[ \frac{x}{\sqrt{9x^2-25}} \right] + C }\)

Problem Statement

Evaluate \(\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } }\) using trig substitution. Give your answer in exact, simplified, factored form.

Solution

PatrickJMT - 82 video solution

video by PatrickJMT

Final Answer

\(\displaystyle{ \int{ \frac{dx}{(9x^2-25)^{3/2}} } = \frac{-1}{25} \left[ \frac{x}{\sqrt{9x^2-25}} \right] + C }\)

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Practice Instructions

Evaluate these integrals using secant or cosecant substitution. Give your answers in exact, simplified and factored form.

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