Integration Of Inverse Trigonometric Functions

Q: What's the difference between integrating $ \arcsin(x) $ and an integral that results in $ \arcsin(x) $?

This lesson focuses on integrals that *result* in $ \arcsin(x) $, like integrating $ \frac{1}{\sqrt{1-x^2}} $. The separate, more advanced process of finding the integral of $ \arcsin(x) $ itself requires a technique called Integration by Parts, which is typically taught later in a calculus course.

Q: Do I always need to use u-substitution?

No, not always. If the variable part of your formula, $ u $, is just a plain $ x $ (so $ u=x $ and $ du=dx $), then you can apply the formula directly. You must use u-substitution whenever the variable part is more complex, like $ u = 4x $ or $ u = x-5 $.

Q: What if the expression in the square root is $ \sqrt{x^2 + a^2} $?

An integral with a denominator like $ \sqrt{x^2 + a^2} $ or $ \sqrt{x^2 - a^2} $ does not match the inverse trigonometric forms taught here. Those forms require a different, more advanced technique called trigonometric substitution.

Q: Why is the absolute value used in the arcsecant formula?

The absolute value in $ \text{arcsec}(\frac{|u|}{a}) $ ensures that the input to the arcsecant function is positive. This is a convention to keep the range of the arcsecant function consistent, typically $ [0, \pi/2) \cup (\pi/2, \pi] $, and aligns with the derivative formula.

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Dive into the world of calculus as we explore the integration of inverse trigonometric functions. This lesson will equip you with the essential formulas and techniques to recognize and solve integrals that result in arcsin, arctan, and arcsec, turning complex problems into manageable steps.

What Is the Integration of Inverse Trigonometric Functions?

The integration of inverse trigonometric functions refers to the process of finding integrals whose results are the inverse trigonometric functions, such as $\arcsin (x)$ , $\arctan (x)$ , or $arcsec (x)$ . This might sound complex, but it's really about pattern recognition. You aren't learning to integrate the function $\arcsin (x)$ itself (that's a more advanced topic called Integration by Parts). Instead, you're learning to integrate expressions that look like the derivatives of inverse trigonometric functions.

Think of it like reversing a process you already know. In differentiation, you found that the derivative of $\arcsin (x)$ is $\frac{1}{\sqrt{1 - x^{2}}}$ . Therefore, the integral of $\frac{1}{\sqrt{1 - x^{2}}}$ must be $\arcsin (x) + C$ . This lesson is all about spotting these special fractional forms and applying the correct formula to find the antiderivative.

Why Do We Need to Know Derivatives First?

The Fundamental Theorem of Calculus establishes a profound link between differentiation and integration: they are inverse operations. To successfully recognize an integral that will result in an inverse trigonometric function, you must first be familiar with what their derivatives look like. Memorizing or at least recognizing these derivative forms is the key to unlocking the integration patterns.

Here is a table of the derivatives of the six inverse trigonometric functions. Notice the similarities and differences, especially between the co-function pairs (like arcsin/arccos).

Function $f (x)$	Derivative $f^{'} (x) = \frac{d y}{d x}$
$\arcsin (u)$	$\frac{u^{'}}{\sqrt{1 - u^{2}}}$
$\arccos (u)$	$- \frac{u^{'}}{\sqrt{1 - u^{2}}}$
$\arctan (u)$	$\frac{u^{'}}{1 + u^{2}}$
$arccot (u)$	$- \frac{u^{'}}{1 + u^{2}}$
$arcsec (u)$	$\frac{u^{'}}{\| u \| \sqrt{u^{2} - 1}}$
$arccsc (u)$	$- \frac{u^{'}}{\| u \| \sqrt{u^{2} - 1}}$

Because the derivatives of $\arccos (u)$ , $arccot (u)$ , and $arccsc (u)$ are just the negatives of their counterparts, we typically only need to memorize the integration rules for $\arcsin$ , $\arctan$ , and $arcsec$ . Any integral that looks like it might result in arccos can be solved with the arcsin rule by factoring out a negative sign.

What Are the Main Integration Formulas?

From the derivatives we just reviewed, we can derive the three fundamental integration formulas. These formulas are generalized to work with any constant, which we'll call $a$ , and any function of $x$ , which we'll call $u$ . Your goal in solving these problems is to make your integral match one of these three forms exactly.

1. The Arcsin Formula

This formula is used when you have a denominator with the square root of a constant squared minus a function squared: $\sqrt{a^{2} - u^{2}}$ .

\int \frac{d u}{\sqrt{a^{2} - u^{2}}} = \arcsin (\frac{u}{a}) + C

2. The Arctan Formula

This formula applies when the denominator is the sum of a constant squared and a function squared, with no square root: $a^{2} + u^{2}$ .

\int \frac{d u}{a^{2} + u^{2}} = \frac{1}{a} \arctan (\frac{u}{a}) + C

3. The Arcsec Formula

This is the most specific pattern. Look for a denominator where a function $u$ is outside a square root containing that same function squared minus a constant squared: $u \sqrt{u^{2} - a^{2}}$ .

\int \frac{d u}{u \sqrt{u^{2} - a^{2}}} = \frac{1}{a} arcsec (\frac{| u |}{a}) + C

In all these formulas, $a$ is a positive constant and $u$ is a function of $x$ . Notice that the arctan and arcsec formulas have a $\frac{1}{a}$ coefficient in front, which is a common point of error if forgotten.

How Do You Solve These Integration Problems?

Solving these integrals is a systematic process of pattern matching and substitution. By following these steps, you can break down any problem into a manageable task.

Analyze the Integrand: Look at the function you need to integrate. Pay close attention to the denominator. Does it have a square root? Is it a sum or difference of squares? This initial analysis will point you toward the correct formula (arcsin, arctan, or arcsec).
Identify $a^{2}$ and $u^{2}$ : Once you've chosen a potential formula, identify the constant part ( $a^{2}$ ) and the variable part ( $u^{2}$ ). From these, determine the values of $a$ and $u$ .
Perform u-Substitution (if needed): If your $u$ is anything more complex than just $x$ (e.g., $u = 3 x$ or $u = x - 1$ ), you must perform a formal u-substitution. Find the differential $d u$ by taking the derivative of $u$ with respect to $x$ (i.e., find $d u = u^{'} d x$ ). Solve for $d x$ and substitute both $u$ and the new expression for $d x$ into the integral.
Apply the Formula: Once your integral is perfectly in the form $\int \frac{d u}{\dots}$ , apply the corresponding arcsin, arctan, or arcsec formula.
Substitute Back and Add C: Replace $u$ and $a$ with their original expressions in terms of $x$ . Finally, never forget to add the constant of integration, $+ C$ , to your final answer.

Example 1

Find the integral $\int \frac{d x}{\sqrt{16 - x^{2}}}$ .

Step 1: Analyze the Integrand. The denominator is $\sqrt{16 - x^{2}}$ . This matches the form $\sqrt{a^{2} - u^{2}}$ , which points to the $\arcsin$ formula.

Step 2: Identify $a^{2}$ and $u^{2}$ .
We can see that $a^{2} = 16$ , so $a = 4$ .
We also have $u^{2} = x^{2}$ , so $u = x$ .

Step 3: Perform u-Substitution. Since $u = x$ , the differential is $d u = 1 \cdot d x$ , or simply $d u = d x$ . No complex substitution is needed here.

Step 4: Apply the Formula. We substitute our values into the arcsin formula: $\int \frac{d u}{\sqrt{a^{2} - u^{2}}} = \arcsin (\frac{u}{a}) + C$ .
This gives us: $\int \frac{d x}{\sqrt{16 - x^{2}}} = \arcsin (\frac{x}{4}) + C$ .

Step 5: Substitute Back and Add C. The answer is already in terms of $x$ and we've included $+ C$ . So, the final answer is $\arcsin (\frac{x}{4}) + C$ .

Can We See More Worked Examples?

Absolutely. The key to mastering these is practice, especially with u-substitution. Let's walk through examples that require a bit more algebraic manipulation.

Example 2

Evaluate the integral $\int \frac{5}{9 + 4 x^{2}} d x$ .

Step 1: Analyze the Integrand. The denominator is $9 + 4 x^{2}$ , a sum of squares with no square root. This is a clear indicator for the $\arctan$ formula. We can pull the constant 5 out of the integral to simplify: $5 \int \frac{d x}{9 + 4 x^{2}}$ .

Step 2: Identify $a^{2}$ and $u^{2}$ .
The constant term is $a^{2} = 9$ , so $a = 3$ .
The variable term is $u^{2} = 4 x^{2}$ , so $u = \sqrt{4 x^{2}} = 2 x$ .

Step 3: Perform u-Substitution. Our $u$ is $2 x$ . We find its differential: $d u = 2 d x$ . To substitute for $d x$ in our integral, we solve for it: $d x = \frac{1}{2} d u$ .

Step 4: Apply the Formula. Now we substitute $u$ and $d x$ into our integral:
$5 \int \frac{d x}{9 + 4 x^{2}} = 5 \int \frac{\frac{1}{2} d u}{a^{2} + u^{2}} = \frac{5}{2} \int \frac{d u}{a^{2} + u^{2}}$
Now the integral perfectly matches the arctan form. We apply the formula $\int \frac{d u}{a^{2} + u^{2}} = \frac{1}{a} \arctan (\frac{u}{a}) + C$ :
$\frac{5}{2} [\frac{1}{3} \arctan (\frac{u}{3})] + C = \frac{5}{6} \arctan (\frac{u}{3}) + C$

Step 5: Substitute Back and Add C. Finally, we replace $u$ with $2 x$ to get our answer in terms of the original variable:
$\frac{5}{6} \arctan (\frac{2 x}{3}) + C$

Example 3

Find the integral $\int \frac{d x}{x \sqrt{25 x^{2} - 4}}$ .

Step 1: Analyze the Integrand. The denominator has the structure $x$ times a square root of a variable part minus a constant part. This strongly suggests the $arcsec$ formula, $\int \frac{d u}{u \sqrt{u^{2} - a^{2}}}$ .

Step 2: Identify $a^{2}$ and $u^{2}$ .
The constant term inside the root is $a^{2} = 4$ , so $a = 2$ .
The variable term is $u^{2} = 25 x^{2}$ , so $u = 5 x$ .

Step 3: Perform u-Substitution. We have $u = 5 x$ , so $d u = 5 d x$ , which means $d x = \frac{1}{5} d u$ . We also need to substitute for the $x$ outside the square root. From $u = 5 x$ , we get $x = \frac{u}{5}$ .

Step 4: Apply the Formula. Let's substitute everything into the integral:
$\int \frac{d x}{x \sqrt{25 x^{2} - 4}} = \int \frac{\frac{1}{5} d u}{(\frac{u}{5}) \sqrt{u^{2} - a^{2}}}$
Notice how the $\frac{1}{5}$ in the numerator and the $\frac{1}{5}$ in the denominator cancel out perfectly! This is a common occurrence in arcsec problems.
$= \int \frac{d u}{u \sqrt{u^{2} - a^{2}}}$
Now we apply the arcsec formula: $\frac{1}{a} arcsec (\frac{| u |}{a}) + C$ .
$\frac{1}{2} arcsec (\frac{| u |}{2}) + C$

Step 5: Substitute Back and Add C. We replace $u$ with $5 x$ for our final answer:
$\frac{1}{2} arcsec (\frac{| 5 x |}{2}) + C$

What If the Denominator Doesn't Match the Formula?

Sometimes, the denominator is a quadratic expression that doesn't immediately look like $a^{2} + u^{2}$ or $a^{2} - u^{2}$ . When you see a quadratic with a middle $x$ term, such as $x^{2} + 6 x + 10$ , your strategy should be to use the algebraic technique of completing the square. This method rewrites the quadratic into a squared term plus or minus a constant, forcing it to match one of our patterns.

The process for completing the square on $x^{2} + b x$ is to add and subtract $(b / 2)^{2}$ . This creates a perfect square trinomial.

Example 4

Evaluate $\int \frac{d x}{x^{2} - 8 x + 25}$ .

Step 1: Analyze the Integrand. The denominator is a quadratic trinomial. It doesn't have a square root, so if it fits any pattern, it will be the $\arctan$ form. We need to manipulate the denominator by completing the square.

Step 2: Complete the Square.
Focus on the $x^{2} - 8 x$ part. The coefficient $b$ is -8.
Calculate $(b / 2)^{2} = (- 8 / 2)^{2} = (- 4)^{2} = 16$ .
Add and subtract 16 within the denominator:
$x^{2} - 8 x + 16 - 16 + 25$
Group the first three terms, which form a perfect square, and combine the constants:
$(x^{2} - 8 x + 16) + (- 16 + 25) = (x - 4)^{2} + 9$
Our integral is now: $\int \frac{d x}{(x - 4)^{2} + 9}$ .

Step 3: Identify $a^{2}$ and $u^{2}$ .
Now it matches the $\arctan$ form $\int \frac{d u}{u^{2} + a^{2}}$ .
The constant term is $a^{2} = 9$ , so $a = 3$ .
The variable term is $u^{2} = (x - 4)^{2}$ , so $u = x - 4$ .

Step 4: Perform u-Substitution.
With $u = x - 4$ , we find the differential $d u = 1 d x$ , so $d u = d x$ .

Step 5: Apply the Formula and Substitute Back.
The integral becomes $\int \frac{d u}{u^{2} + a^{2}}$ .
Applying the arctan formula gives: $\frac{1}{a} \arctan (\frac{u}{a}) + C$ .
Substituting our values for $u$ and $a$ :
$\frac{1}{3} \arctan (\frac{x - 4}{3}) + C$

What Are Common Mistakes to Avoid?

When working through these problems, students often fall into a few common traps. Being aware of these can help you double-check your work and improve your accuracy.

Forgetting the $\frac{1}{a}$ Coefficient: This is the most frequent error. The formulas for $\arctan$ and $arcsec$ both have a $\frac{1}{a}$ multiplier in the final answer. The $\arcsin$ formula does not. It's easy to forget this factor in the heat of solving a problem.
Mixing Up Arcsin and Arctan Forms: Students sometimes confuse the denominators. Remember: $\sqrt{a^{2} - u^{2}}$ (square root of a difference) leads to arcsin. $a^{2} + u^{2}$ (a sum, no root) leads to arctan.
Incorrect u-Substitution: A failed u-substitution can derail the entire problem. Always remember that when you substitute $u$ , you must also substitute $d u$ . Forgetting to replace $d x$ with its equivalent in terms of $d u$ is a critical mistake.
Errors in Completing the Square: The algebra of completing the square can be tricky. Be careful when calculating $(b / 2)^{2}$ and ensure you both add and subtract it to keep the expression equivalent.
Forgetting $+ C$ : Every indefinite integral must have the constant of integration, $+ C$ , appended to the final answer. It represents the family of all possible antiderivative functions. Leaving it off will almost always result in a loss of points.

Quick Reference: Key Formulas

For quick review and memorization, here are the three core formulas for integrating functions that result in inverse trigonometric functions. Your goal is to use algebraic manipulation and u-substitution to make your problem match one of these templates.

Resulting Function	Integration Formula	Key Denominator Pattern
$\arcsin$	$\int \frac{d u}{\sqrt{a^{2} - u^{2}}} = \arcsin (\frac{u}{a}) + C$	Square root of (constant - variable)
$\arctan$	$\int \frac{d u}{a^{2} + u^{2}} = \frac{1}{a} \arctan (\frac{u}{a}) + C$	Sum of squares, no root
$arcsec$	$\int \frac{d u}{u \sqrt{u^{2} - a^{2}}} = \frac{1}{a} arcsec (\frac{\| u \|}{a}) + C$	Variable times root of (variable - constant)

Frequently Asked Questions

Why are there only three main integration formulas if there are six inverse trig functions?

The other three inverse functions (arccosine, arccotangent, arcsecant) have derivatives that are just the negative versions of the main three. Since we can always factor a constant (-1) out of an integral, we don't need separate rules for them. For example, an integral that looks like it should be arccos can be solved using the arcsin rule and multiplying by -1.

What's the difference between integrating

\arcsin (x)

and an integral that results in

\arcsin (x)

This lesson focuses on integrals that *result* in $\arcsin (x)$ , like integrating $\frac{1}{\sqrt{1 - x^{2}}}$ . The separate, more advanced process of finding the integral of $\arcsin (x)$ itself requires a technique called Integration by Parts, which is typically taught later in a calculus course.

How can I tell which inverse trig formula to use?

Look at the structure of the denominator. If it's a square root of a constant minus a variable part, think arcsin. If it's a sum of squares with no square root, think arctan. If it's a variable outside a square root of a variable part minus a constant, think arcsec.

Do I always need to use u-substitution?

No, not always. If the variable part of your formula, $u$ , is just a plain $x$ (so $u = x$ and $d u = d x$ ), then you can apply the formula directly. You must use u-substitution whenever the variable part is more complex, like $u = 4 x$ or $u = x - 5$ .

\arctan (x)

the same as

\tan^{- 1} (x)

Yes, they are two different notations for the exact same function: the inverse tangent. The $\arctan (x)$ notation is often preferred to avoid confusion with the reciprocal, $(\tan (x))^{- 1} = \frac{1}{\tan (x)} = \cot (x)$ .

What if the expression in the square root is

\sqrt{x^{2} + a^{2}}

An integral with a denominator like $\sqrt{x^{2} + a^{2}}$ or $\sqrt{x^{2} - a^{2}}$ does not match the inverse trigonometric forms taught here. Those forms require a different, more advanced technique called trigonometric substitution.

Why is the absolute value used in the arcsecant formula?

The absolute value in $arcsec (\frac{| u |}{a})$ ensures that the input to the arcsecant function is positive. This is a convention to keep the range of the arcsecant function consistent, typically $[0, π / 2) \cup (π / 2, π]$ , and aligns with the derivative formula.