Integrals Involving Inverse Trigonometric Functions: Videos & Practice Problems

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Learn to integrate functions resulting in inverse trigonometric functions by reversing derivative rules. For example, the integral of $\frac{1}{\sqrt{1-x^2}}$ is the inverse sine of x. Similarly, integrate $\frac{1}{1+x^2}$ to get the inverse tangent of x. Use substitution for integrals like $\frac{1}{\sqrt{a^2-u^2}}$ to find inverse sine of $\frac{u}{a}$. Practice these rules for mastery.

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Integrals Resulting in Inverse Trig Functions

In this section, we explore the integration of certain functions related to inverse trigonometric functions, leveraging the relationship between derivatives and integrals. Understanding these integrals is crucial, as they often arise in calculus problems where standard integration techniques may not suffice.

We begin with the integral of the function 1 / √(1 - x²). Recognizing that the derivative of the inverse sine function, sin^-1(x), is 1 / √(1 - x²), we can conclude that:

$$\int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1}(x) + C$$

Here, C represents the constant of integration. For example, when integrating 4 / √(1 - x²), we can factor out the constant:

$$\int \frac{4}{\sqrt{1 - x^2}} \, dx = 4 \cdot \sin^{-1}(x) + C$$

Next, we consider the integral of 1 / (1 + x²). The derivative of the inverse tangent function, tan^-1(x), is 1 / (1 + x²), leading to the integral:

$$\int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) + C$$

For a more complex example, such as 1 / (1 + x²) + 1 / x², we can integrate each term separately. The first term gives us tan^-1(x), while the second term can be rewritten as x^-2, allowing us to apply the power rule:

$$\int \frac{1}{x^2} \, dx = -\frac{1}{x} + C$$

Thus, the combined result is:

$$\int \left(\frac{1}{1 + x^2} + \frac{1}{x^2}\right) \, dx = \tan^{-1}(x) - \frac{1}{x} + C$$

Finally, we examine the integral of 1 / (|x|√(x² - 1)). The derivative of the inverse secant function, sec^-1(x), is 1 / (|x|√(x² - 1)), leading to:

$$\int \frac{1}{|x| \sqrt{x^2 - 1}} \, dx = \sec^{-1}(x) + C$$

For an example involving a constant, such as 1 / (3|x|√(x² - 1)), we can factor out the constant:

$$\int \frac{1}{3|x| \sqrt{x^2 - 1}} \, dx = \frac{1}{3} \sec^{-1}(x) + C$$

In summary, recognizing the derivatives of inverse trigonometric functions allows us to derive their corresponding integrals effectively. Memorizing these integral rules is beneficial, especially when encountering functions with 1 and x² in the denominator, as they often indicate the use of inverse trig functions. Continued practice with these integrals will enhance your proficiency in calculus.

Evaluate the integral.
${\int }_{}^{}(\frac{2}{\sqrt{1-x^2}}-\frac{1}{3\sqrt{x}})dx$

• A. $2co{s}^{-1}x-\frac{2}{3}\sqrt{x}+C$
• B. $2si{n}^{-1}x-\frac{2}{3}\sqrt{x}+C$
• C. $co{s}^{-1}x-\frac{2}{3}\sqrt{x}+C$
• D. $si{n}^{-1}x-\frac{2}{3}\sqrt{x}+C$

Answer: (B)

Integrals Resulting in Inverse Trig Functions Example 1

In this section, we explore the integration of certain functions related to inverse trigonometric functions, leveraging the relationship between derivatives and integrals. Understanding these integrals can significantly enhance our ability to solve complex problems that may not be easily addressed with standard integration techniques.

We begin with the integral of the function 1 / √(1 - x²). Recognizing that the derivative of the inverse sine function, sin^-1(x), is 1 / √(1 - x²), we can conclude that:

$$\int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1}(x) + C$$

In an example, when integrating 4 / √(1 - x²), we can factor out the constant:

$$\int \frac{4}{\sqrt{1 - x^2}} \, dx = 4 \int \frac{1}{\sqrt{1 - x^2}} \, dx = 4 \sin^{-1}(x) + C$$

Next, we consider the integral of 1 / (1 + x²). The derivative of the inverse tangent function, tan^-1(x), is 1 / (1 + x²), leading us to the integral:

$$\int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) + C$$

For a more complex example, integrating 1 / (1 + x²) + 1 / x² involves breaking it into two parts. The first part directly applies our previous rule, while the second part can be rewritten using the power rule:

$$\int \left(\frac{1}{1 + x^2} + \frac{1}{x^2}\right) \, dx = \tan^{-1}(x) - \frac{1}{x} + C$$

Finally, we examine the integral of 1 / (|x|√(x² - 1)). The derivative of the inverse secant function, sec^-1(x), is 1 / (|x|√(x² - 1)), allowing us to express the integral as:

$$\int \frac{1}{|x| \sqrt{x^2 - 1}} \, dx = \sec^{-1}(x) + C$$

In an example involving 1 / (3|x|√(x² - 1)), we can factor out the constant:

$$\int \frac{1}{3|x| \sqrt{x^2 - 1}} \, dx = \frac{1}{3} \sec^{-1}(x) + C$$

As we work with these integral rules, it is beneficial to memorize them, especially when encountering expressions with 1 and x² in the denominator, as they often indicate the use of inverse trigonometric functions. Continued practice with these integrals will solidify your understanding and application of these concepts.

Evaluate the integral.
${\int }_{\sqrt{2}}^2\frac{1}{\mid x\mid \sqrt{x^2-1}}dx$

• A. $\frac{\pi }{12}$
• B. $-\frac{\pi }{12}$
• C. $\frac{7\pi }{12}$
• D. $\frac{5\pi }{12}$

Answer: (A)

Integrals Resulting in Inverse Trig Functions: Substitution

In the study of integrals, we encounter various forms that can lead to inverse trigonometric functions. A key rule to remember is the integral of the form:

$$\int \frac{1}{\sqrt{a^2 - u^2}} \, du = \arcsin\left(\frac{u}{a}\right) + C$$

Here, \(a\) is a positive constant, and \(u\) is a function of \(x\). This integral results in the inverse sine function, and if we set \(u = x\) and \(a = 1\), it simplifies to the familiar inverse sine of \(x\).

For example, consider the integral:

$$\int \frac{1}{\sqrt{16 - 9x^2}} \, dx$$

We can rewrite this as:

$$\int \frac{1}{\sqrt{4^2 - (3x)^2}} \, dx$$

Here, \(a = 4\) and we make the substitution \(u = 3x\), leading to \(du = 3 \, dx\) or \(dx = \frac{du}{3}\). Adjusting for this substitution, we have:

$$\int \frac{1}{\sqrt{4^2 - u^2}} \cdot \frac{du}{3} = \frac{1}{3} \arcsin\left(\frac{u}{4}\right) + C$$

Substituting back gives us:

$$\frac{1}{3} \arcsin\left(\frac{3x}{4}\right) + C$$

Next, we explore the integral:

$$\int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$$

For instance, integrating:

$$\int \frac{6}{25 + 36x^2} \, dx$$

We rewrite this as:

$$\int \frac{6}{5^2 + (6x)^2} \, dx$$

With \(a = 5\) and \(u = 6x\), we find \(du = 6 \, dx\), allowing us to express \(dx\) as \(\frac{du}{6}\). Thus, we have:

$$\int \frac{6}{5^2 + u^2} \cdot \frac{du}{6} = \frac{1}{5} \arctan\left(\frac{u}{5}\right) + C$$

Substituting back yields:

$$\frac{1}{5} \arctan\left(\frac{6x}{5}\right) + C$$

Finally, we consider the integral:

$$\int \frac{1}{|u| \sqrt{u^2 - a^2}} \, du = \frac{1}{a} \operatorname{arcsec}\left(\frac{|u|}{a}\right) + C$$

For example, integrating:

$$\int \frac{7}{|7x| \sqrt{49x^2 - 4}} \, dx$$

We rewrite this as:

$$\int \frac{7}{|7x| \sqrt{(7x)^2 - 2^2}} \, dx$$

With \(u = 7x\) and \(a = 2\), we find \(du = 7 \, dx\), leading to:

$$\int \frac{7}{|u| \sqrt{u^2 - 2^2}} \cdot \frac{du}{7} = \frac{1}{2} \operatorname{arcsec}\left(\frac{|u|}{2}\right) + C$$

Substituting back gives us:

$$\frac{1}{2} \operatorname{arcsec}\left(\frac{|7x|}{2}\right) + C$$

These integral rules, combined with substitution techniques, provide a robust framework for solving integrals that yield inverse trigonometric functions. Mastery of these concepts is best achieved through practice and application in various problems.

Integrals Resulting in Inverse Trig Functions: Substitution Example 2

To evaluate the integral of the function \( \frac{e^x}{4 + e^{2x}} \, dx \), we can simplify the expression by rewriting it. Notice that \( 4 + e^{2x} \) can be expressed as \( 2^2 + (e^x)^2 \). This allows us to identify the integral in the form of \( \frac{1}{a^2 + u^2} \), where \( a = 2 \) and \( u = e^x \).

Next, we perform a substitution where \( u = e^x \). Consequently, the differential \( du \) becomes \( e^x \, dx \), which matches the numerator of our integral. This substitution transforms our integral into:

\[\int \frac{du}{4 + u^2} = \int \frac{du}{2^2 + u^2}\]

Using the integral formula for \( \int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \), we can substitute \( a = 2 \) into the equation. Thus, we have:

\[\int \frac{du}{2^2 + u^2} = \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) + C\]

Substituting back \( u = e^x \), the integral evaluates to:

\[\frac{1}{2} \tan^{-1} \left( \frac{e^x}{2} \right) + C\]

In conclusion, the final result for the integral \( \int \frac{e^x}{4 + e^{2x}} \, dx \) is:

\[\frac{1}{2} \tan^{-1} \left( \frac{e^x}{2} \right) + C\]

Find the indefinite integral.
${\int }_{}^{}\frac{1}{x\sqrt{4x^2-16}}dx$

• A. $\frac{1}{4}cs{c}^{-1}\mid \frac{x}{2}\mid +C$
• B. $se{c}^{-1}\mid \frac{x}{2}\mid +C$
• C. $\frac{1}{4}se{c}^{-1}\mid \frac{x}{2}\mid +C$
• D. $\frac{1}{4}se{c}^{-1}\mid \frac{x}{4}\mid +C$

Answer: (C)

Evaluate the definite integral in terms of an inverse trig function.
$\int_0^{\frac{\sqrt3}{3}}\frac{dx}{\sqrt{4-9x^2}}$

• A. $\frac{\pi}{3}$
• B. $-\frac{\pi}{3}$
• C. $\frac{\pi }{9}$
• D. $-\frac{\pi}{9}$

Answer: (C)

Evaluate the definite integral using the appropriate substitutions.
${\int }_{\frac{\pi }{2}}^{\pi }\frac{\sin x}{1+co{s}^{2}x}dx$

• A. 0
• B. $\frac{\pi }{4}$
• C. $-\frac{\pi }{4}$
• D. $\frac{\pi }{8}$

Answer: (B)

Integrals Resulting in Inverse Trig Functions: Substitution Example 3

To find the indefinite integral of \( e^{\tan^{-1}(\theta)} \) over \( 1 + \theta^2 \, d\theta \), we can simplify the process using a substitution method. The integral can initially appear complex, but recognizing the structure allows for a straightforward approach.

We start by letting \( u = \tan^{-1}(\theta) \). The derivative of this function, \( du \), is given by:

\[du = \frac{1}{1 + \theta^2} \, d\theta\]

This relationship is crucial because it matches the integrand's structure. Substituting \( u \) into the integral transforms it into:

\[\int e^u \, du\]

Now, integrating \( e^u \) yields:

\[e^u + C\]

Finally, we substitute back \( u = \tan^{-1}(\theta) \) to express the result in terms of \( \theta \):

\[e^{\tan^{-1}(\theta)} + C\]

This process illustrates the power of substitution in simplifying integrals, especially when the integrand contains a function and its derivative. Understanding these relationships enhances your ability to tackle more complex integrals effectively.

Find the indefinite integral.
${\int }_{}^{}\frac{arcsi{n}^{4}x}{\sqrt{1-x^2}}dx$

• A. $arcsi{n}^{5}x+C$
• B. $\frac{1}{3}arcsi{n}^{3}x+C$
• C. $-\frac{1}{5}arcco{s}^{5}x+C$
• D. $\frac{1}{5}arcsi{n}^{5}x+C$

Answer: (D)

Find the indefinite integral.
${\int }_{}^{}\frac{se{c}^2\left(si{n}^{-1}\theta \right)}{\sqrt{1-{\theta }^2}}d\theta$

• A.
  $\tan \left(si{n}^{-1}\theta \right)+C$
• B. $\tan \left(co{s}^{-1}\theta \right)+C$
• C. $\cot \left(si{n}^{-1}\theta \right)+C$
• D. $si{n}^{-1}\left(tan\theta \right)+C$

Answer: (A)

Integrals Resulting in Inverse Trig Functions: Substitution Example 4

To find the indefinite integral of the function \(\frac{t + 5}{t^2 + 16}\,dt\), we can split the integral into two separate fractions:

\[\int \frac{t + 5}{t^2 + 16}\,dt = \int \frac{t}{t^2 + 16}\,dt + \int \frac{5}{t^2 + 16}\,dt\]

For the first integral, \(\int \frac{t}{t^2 + 16}\,dt\), we can use the substitution method. Let \(u = t^2 + 16\). Then, the differential \(du = 2t\,dt\), which implies \(dt = \frac{du}{2t}\). To adjust for the constant, we multiply by \(\frac{1}{2}\) and rewrite the integral:

\[\int \frac{t}{t^2 + 16}\,dt = \frac{1}{2} \int \frac{1}{u}\,du\]

This integral evaluates to:

\[\frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |t^2 + 16| + C\]

Next, we consider the second integral, \(\int \frac{5}{t^2 + 16}\,dt\). We can factor out the constant 5:

\[5 \int \frac{1}{t^2 + 16}\,dt\]

We can rewrite \(16\) as \(4^2\), leading to:

\[5 \int \frac{1}{4^2 + t^2}\,dt\]

This matches the form \(\int \frac{1}{a^2 + u^2}\,du\), where \(a = 4\) and \(u = t\). The result of this integral is:

\[\frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C = \frac{1}{4} \tan^{-1}\left(\frac{t}{4}\right) + C\]

Thus, the second integral evaluates to:

\[5 \cdot \frac{1}{4} \tan^{-1}\left(\frac{t}{4}\right) = \frac{5}{4} \tan^{-1}\left(\frac{t}{4}\right)\]

Now, combining both integrals, we have:

\[\int \frac{t + 5}{t^2 + 16}\,dt = \frac{1}{2} \ln(t^2 + 16) + \frac{5}{4} \tan^{-1}\left(\frac{t}{4}\right) + C\]

In conclusion, the solution to the indefinite integral is:

\[\frac{1}{2} \ln(t^2 + 16) + \frac{5}{4} \tan^{-1}\left(\frac{t}{4}\right) + C\]

Integrals Resulting in Inverse Trig Functions: Substitution Example 5

To compute the definite integral from \(\frac{\sqrt{3}}{3}\) to \(1\) of \(\frac{2}{\arctan(x)(1+x^2)} \, dx\), we start by finding the antiderivative of the integrand. Recognizing the presence of \(\arctan(x)\) and \(1+x^2\), we can use substitution. Let \(u = \arctan(x)\), which gives us \(du = \frac{1}{1+x^2} \, dx\). This allows us to rewrite the integral as:

\[\int \frac{2}{\arctan(x)(1+x^2)} \, dx = 2 \int \frac{1}{u} \, du\]

Using the integral of \(\frac{1}{u}\), we find:

\[2 \int \frac{1}{u} \, du = 2 \ln |u| + C = 2 \ln |\arctan(x)| + C\]

Since we are dealing with a definite integral, we will evaluate this from \(\frac{\sqrt{3}}{3}\) to \(1\). Thus, we have:

\[2 \left( \ln |\arctan(1)| - \ln |\arctan(\frac{\sqrt{3}}{3})| \right)\]

Calculating the values of the arctangent, we find \(\arctan(1) = \frac{\pi}{4}\) and \(\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}\). Substituting these values gives:

\[2 \left( \ln \left(\frac{\pi}{4}\right) - \ln \left(\frac{\pi}{6}\right) \right)\]

Using the property of logarithms that states \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\), we can simplify this to:

\[2 \ln \left(\frac{\frac{\pi}{4}}{\frac{\pi}{6}}\right) = 2 \ln \left(\frac{6}{4}\right) = 2 \ln \left(\frac{3}{2}\right)\]

This expression represents the exact value of the definite integral. If a numerical approximation is desired, evaluating \(2 \ln \left(\frac{3}{2}\right)\) yields approximately \(0.81\).

Completing the Square to Rewrite the Integrand

In this lesson, we explore the process of evaluating integrals that may not initially appear to fit standard forms, particularly those leading to inverse trigonometric functions. A common example is the integral of the form:

$$\int \frac{7}{x^2 + 6x + 13} \, dx$$

To tackle this integral, we first need to rewrite the denominator into a recognizable form suitable for applying inverse trigonometric identities. This often involves completing the square, a technique that transforms a quadratic expression into a perfect square trinomial.

Starting with the quadratic expression \(x^2 + 6x + 13\), we identify the coefficient \(b = 6\). To complete the square, we calculate \(b/2\) and square it:

$$\left(\frac{6}{2}\right)^2 = 3^2 = 9$$

We then rewrite the expression by adding and subtracting this value:

$$x^2 + 6x + 9 + 4 = (x + 3)^2 + 4$$

This allows us to express the integral as:

$$\int \frac{7}{(x + 3)^2 + 4} \, dx$$

Next, we can factor out the constant 7, leading to:

$$7 \int \frac{1}{(x + 3)^2 + 2^2} \, dx$$

This integral now resembles the standard form for the inverse tangent function:

$$\int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$$

In our case, we set \(u = x + 3\) and \(a = 2\). The differential \(du\) is simply \(dx\). Thus, we can evaluate the integral:

$$7 \cdot \frac{1}{2} \tan^{-1}\left(\frac{x + 3}{2}\right) + C$$

Combining these results gives us the final answer:

$$\frac{7}{2} \tan^{-1}\left(\frac{x + 3}{2}\right) + C$$

In summary, when faced with integrals that do not conform to standard rules, completing the square is a valuable technique. It allows us to rewrite the integrand into a form that can be evaluated using known inverse trigonometric identities. This method is particularly useful when other techniques, such as the power rule or basic substitutions, do not yield results.

Completing the Square to Rewrite the Integrand Example 6

To find the indefinite integral of the function \(\frac{8}{\sqrt{4x - x^2}} \, dx\), we first need to rewrite the expression under the square root in a more manageable form. This involves completing the square for the quadratic expression \(4x - x^2\).

We start by rearranging the quadratic into standard form: \(-x^2 + 4x\). To complete the square, we factor out a negative sign, yielding \(- (x^2 - 4x)\). Next, we add and subtract the square of half the coefficient of \(x\) (which is \(-2\)) inside the parentheses. This gives us:

\[-(x^2 - 4x + 4 - 4) = -((x - 2)^2 - 4) = - (x - 2)^2 + 4.\]

Thus, we can rewrite the original integral as:

\[\int \frac{8}{\sqrt{4 - (x - 2)^2}} \, dx.\]

Recognizing this form, we can relate it to the standard integral for inverse sine, which is:

\[\int \frac{1}{\sqrt{a^2 - u^2}} \, du = \arcsin\left(\frac{u}{a}\right) + C,\]

where \(a = 2\) and \(u = x - 2\). The differential \(du\) is simply \(dx\), allowing us to rewrite the integral as:

\[8 \int \frac{1}{\sqrt{2^2 - u^2}} \, du.\]

Applying the inverse sine formula, we find:

\[8 \left( \arcsin\left(\frac{u}{2}\right) + C \right).\]

Substituting back for \(u\), we arrive at the final result:

\[8 \arcsin\left(\frac{x - 2}{2}\right) + C.\]

This expression represents the indefinite integral of the original function, showcasing the application of completing the square and recognizing the form of the integral related to inverse trigonometric functions.

Find the indefinite integral.
${\int }_{}^{}\frac{-2}{x^2-12x+45}dx$

• A. $-2\ln \mid x^2-12x+45\mid +C$
• B. $2\ln \mid x^2-12x+45\mid +C$
• C. $\frac{2}{3}ta{n}^{-1}\left(\frac{x-6}{3}\right)+C$
• D. $-\frac{2}{3}ta{n}^{-1}\left(\frac{x-6}{3}\right)+C$

Answer: (D)

Completing the Square to Rewrite the Integrand Example 7

When evaluating integrals, it's essential to match each integral with the appropriate method for simplification and calculation. Let's explore several integrals and the strategies to tackle them effectively.

The first integral is $\int \frac{x^2 + 3x - 2}{x - 1} \, dx$. Here, the degree of the polynomial in the numerator is higher than that in the denominator. This suggests using long division to simplify the expression into a more manageable form before integrating.

Next, consider the integral $\int \frac{x + 4}{x^2 + 8x + 5} \, dx$. In this case, the denominator has a higher degree than the numerator. A common approach is to use u-substitution, where we let $u = x^2 + 8x + 5$. The derivative $du = (2x + 8) \, dx$ can be manipulated to facilitate integration, leading to the form $\frac{1}{2} \int \frac{du}{u}$, which is straightforward to evaluate.

For the integral $\int \frac{2x - 3}{\sqrt{x^2 - 3x}} \, dx$, we can set $u = x^2 - 3x$. This substitution allows us to express the integral in terms of $u$, leading to a form that can be integrated using the power rule, specifically $\int u^{-1/2} \, du$.

Next, we have the integral $\int \frac{5}{\sqrt{4 + 3x - x^2}} \, dx$. The absence of $x$ in the numerator suggests a more complex approach. Here, completing the square in the expression under the radical can help transform it into a form suitable for an inverse trigonometric function.

Finally, consider the integral $\int \frac{4}{16x^2 + 9} \, dx$. Similar to the previous case, the quadratic in the denominator and the lack of a variable in the numerator indicate the use of an inverse trigonometric function. By letting $u = 4x$, we can rewrite the integral in a form that aligns with the standard inverse trig formulas.

Understanding these strategies not only aids in solving integrals but also enhances overall mathematical proficiency. Each method—long division, u-substitution, completing the square, and recognizing inverse trig forms—plays a crucial role in simplifying and evaluating integrals effectively.

Here’s what students ask on this topic:

What is the integral of 1 over the square root of 1 minus x squared?

The integral of $\frac{1}{\sqrt{1-x^2}}$ with respect to x is the inverse sine function, denoted as $\sin -1\left(x\right)$, plus a constant of integration, C. This is because the derivative of the inverse sine function is $\frac{1}{\sqrt{1-x^2}}$. Therefore, integrating this expression reverses the differentiation process, resulting in the inverse sine function.

How do you integrate functions that result in inverse trigonometric functions?

To integrate functions that result in inverse trigonometric functions, recognize the form of the integrand. For example, the integral of $\frac{1}{\sqrt{a^2-u^2}}$ results in $\sin -1\left(\frac{u}{a}\right)$ plus C. Similarly, $\frac{1}{a^2}$ results in $\frac{1}{a}\tan -1\left(\frac{u}{a}\right)$ plus C. Use substitution if necessary to match these forms, and remember to adjust for constants.

What is the role of completing the square in evaluating integrals involving inverse trigonometric functions?

Completing the square is a technique used to transform quadratic expressions into a form that matches the integrands of inverse trigonometric functions. For example, if you have an integral like $\frac{1}{x^2}$, completing the square can rewrite the denominator as $(x+d{)}^2+e$, which may resemble the form needed for inverse tangent integration. This process allows you to apply the appropriate inverse trigonometric integral formula, facilitating the evaluation of the integral.

How do you apply substitution in integrals involving inverse trigonometric functions?

Substitution in integrals involving inverse trigonometric functions involves replacing a part of the integrand with a new variable to simplify the integral. For instance, if you have an integral like $\frac{1}{\sqrt{a^2-u^2}}$, you can let $u=g\left(x\right)$ and find $\mathrm{du}$ in terms of $\mathrm{dx}$. This substitution transforms the integral into a standard form, allowing you to apply the inverse trigonometric integral formula. After integration, substitute back to express the result in terms of the original variable.

What is the integral of 1 over the absolute value of x times the square root of x squared minus 1?

The integral of $\frac{1}{\left|}$ with respect to x is the inverse secant function, denoted as $\sec -1\left(x\right)$, plus a constant of integration, C. This is because the derivative of the inverse secant function is $\frac{1}{\left|}$. Therefore, integrating this expression reverses the differentiation process, resulting in the inverse secant function.