📚

 > 

♾️ 

 > 

🔥

6.14 Selecting Techniques for Antidifferentiation (AB)

6 min readjune 18, 2024

6.14 Selecting Techniques for Antidifferentiation

In AP Calculus, we encounter a variety of functions, and each may require a different approach for antidifferentiation. This skill is all about choosing the right technique or method to find the antiderivative of a given expression. Let's explore the techniques necessary for this task in detail! 🚀

🔋Power Rule for Antiderivatives

The power rule for antiderivatives is a fundamental technique used to find the antiderivative of a function raised to a power. When you have an integral in the form ∫xndxx^n dx, this rule is your go-to method. It states that if you havexndx\int x^n dx, where n is not equal to -1, you add 1 to the exponent and divide by the new exponent. This results in:

xndx=xn+1(n+1)+C\int x^n dx=\frac{x^{n+1}}{(n+1)} + C

Where C is the constant of integration.

🚇 U-substitution (U-sub)

U-substitution is a powerful technique for simplifying complex integrals. It involves substituting a portion of the expression with a single variable (usually denoted as 'u') to make the integral more manageable.

This method is particularly useful when dealing with composite functions or expressions that involve chains of functions. The basic steps for u-substitution are as follows:

  1. Choose a suitable 'u' that simplifies the integral.
  2. Find du (the derivative of 'u') and express dx in terms of du.
  3. Rewrite the integral in terms of 'u.'
  4. Integrate with respect to 'u.'
  5. Substitute back in terms of 'x' if necessary.

👉 Trigonometric Functions

Trigonometric functions, like sine, cosine, and tangent, frequently appear in integrals. Understanding how to handle these functions is crucial. For example, when integrating trigonometric expressions, you may use trigonometric identities or specific trigonometric integrals.

Common trigonometric integrals include:

  • sin(x)dx=cos(x)+C∫\sin(x) dx = -\cos(x) + C
  • cos(x)dx=sin(x)+C∫\cos(x) dx = \sin(x) + C
  • tan(x)dx=lncos(x)+C∫\tan(x) dx = -\ln|\cos(x)| + C
  • cot(x)dx=lnsin(x)+C∫\cot(x) dx = \ln|\sin(x)| + C
  • sec(x)dx=lnsec(x)+tan(x)+C∫\sec(x) dx = \ln|\sec(x)+\tan(x)| + C
  • csc(x)dx=lncsc(x)+cot(x)+C∫\csc(x) dx = -\ln|\csc(x)+\cot(x)| + C

🙃 Inverse Trigonometric Functions

Antidifferentiation often involves inverse trigonometric functions like arcsin, arccos, and arctan. Knowing the antiderivatives of these functions and how to apply them is essential.

Here are some common inverse trigonometric integrals:

  • ddx(sin1(x))=11x2\frac {d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}
  • ddx(cos1(x))=11x2\frac {d}{dx}(\cos^{-1}(x)) = \frac{-1}{\sqrt{1-x^2}}
  • ddx(tan1(x))=11+x2\frac {d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2}
  • ddx(cot1(x))=11+x2\frac {d}{dx}(\cot^{-1}(x)) = \frac{-1}{1+x^2}
  • ddx(sec1(x))=1x1x2\frac {d}{dx}(\sec^{-1}(x)) = \frac{1}{x\sqrt{1-x^2}}
  • ddx(csc1(x))=1x1x2\frac {d}{dx}(\csc^{-1}(x)) = \frac{-1}{x\sqrt{1-x^2}}

🪴 Exponentials and Logarithms

Functions involving exponentials (exe^x) and natural logarithms (ln(x)\ln(x)) frequently appear in calculus problems. Familiarity with their antiderivatives is necessary for solving such integrals.

1.exdx=ex+C1. ∫e^x dx = e^x + C
2.1xdx=lnx+C2. ∫\frac{1}{x} dx = \ln| x | + C

Long Division

Long division can be useful when dealing with rational functions, where the degree of the numerator is equal to or higher than the degree of the denominator. It helps simplify the expression before antidifferentiation.

Here is the long division method for the expression:

(x2+2x3)/(x1) dx∫(x^2 + 2x - 3)/(x - 1) \ dx

🥇 Step 1: Perform Long Division

Perform long division to divide the numerator by the denominator:

x2+2x3x1=x3\frac{x^2+2x-3}{x-1} = x-3

🥈 Step 2: Write the Integral as a Sum Rewrite the integral as a sum of two integrals:

(x+3)dx(3(x1))dx∫(x + 3) dx - ∫(\frac{3}{(x - 1)}) dx

🥉 Step 3: Integrate Each Term Separately Now, integrate each term separately:

  • (x+3)dx=12x2+3x+C∫(x + 3) dx = \frac{1}{2}x^2 + 3x + C
  • (3(x1))dx=3lnx1+C∫(\frac{3}{(x - 1)}) dx = 3ln| x - 1 | + C

Step 4: Combine the Results Combine the results to find the antiderivative of the original expression:

(12)x2+3x3lnx1+C(\frac{1}{2})x^2 + 3x - 3ln| x - 1 | + C

🧊 Completing the Square

Completing the square is a technique often used with quadratic functions. It's crucial when you encounter expressions that can be transformed into a perfect square trinomial for easier integration.

Here's a step-by-step example of integrating a quadratic expression by completing the square, for…

(x2+4x+4)dx∫(x^2 + 4x + 4) dx

🥇 Step 1: Identify the Perfect Square Trinomial Recognize that the expression can be factored as a perfect square trinomial: (x+2)2(x + 2)^2.

🥈 Step 2: Rewrite the Integral Rewrite the integral using the perfect square trinomial:

(x2+4x+4)dx=(x+2)2dx∫(x^2 + 4x + 4) dx = ∫(x + 2)^2 dx

🥉 Step 3: Use the Power Rule for Antiderivatives Apply the power rule for antiderivatives to find the integral:

(x+2)2dx=(x+2)33+C∫(x + 2)^2 dx = \frac{(x + 2)^3}{3} + C

Step 4: Simplify Simplify the expression, resulting in its antiderivative:

(x+2)33+C\frac{(x + 2)^3}{3} + C

✏️ Integrating Practice Question

Let’s put those newly learned skills to good use! Find the antiderivative of the following expression:

(ex+2cos(x))dx∫(e^x + 2\cos(x)) dx

Be sure to try the question out yourself before moving onto the solution below!

✅ Solution to Practice Problem

When approaching this problem it is best to first recognize the terms in the expression and determine which technique to use for each term.

  • Term 1: exe^x (Antiderivative of exe^x is exe^x)
  • Term 2: 2cos(x)2\cos(x) (Antiderivative of cos(x)\cos(x) is sin(x)\sin(x))

Now, take the integral of each term individually.

  • Since the integral of exe^x is itself, term 1 is easy! exe^x
  • For term 2, the integral is 2sin(x)2\sin(x).

You got the two separate integrals! All that is left is to combine the results and add the +C+C term to the end.

(ex+2cos(x)) dx=ex+2sin(x)+C∫(e^x+2\cos(x))\ dx = e^x +2\sin(x)+C

So, the antiderivative of the given expression is ex+2sin(x)+Ce^x +2\sin(x)+C. Great work! 👏


⭐ Closing

Guess what?! You made it to the end of unit 6 in AP Calculus! Ready to move on to unit seven? It’s all about differential equations. Check it out here!

image.gif

Image Courtesy of Giphy

<< Hide Menu

📚

 > 

♾️ 

 > 

🔥

6.14 Selecting Techniques for Antidifferentiation (AB)

6 min readjune 18, 2024

6.14 Selecting Techniques for Antidifferentiation

In AP Calculus, we encounter a variety of functions, and each may require a different approach for antidifferentiation. This skill is all about choosing the right technique or method to find the antiderivative of a given expression. Let's explore the techniques necessary for this task in detail! 🚀

🔋Power Rule for Antiderivatives

The power rule for antiderivatives is a fundamental technique used to find the antiderivative of a function raised to a power. When you have an integral in the form ∫xndxx^n dx, this rule is your go-to method. It states that if you havexndx\int x^n dx, where n is not equal to -1, you add 1 to the exponent and divide by the new exponent. This results in:

xndx=xn+1(n+1)+C\int x^n dx=\frac{x^{n+1}}{(n+1)} + C

Where C is the constant of integration.

🚇 U-substitution (U-sub)

U-substitution is a powerful technique for simplifying complex integrals. It involves substituting a portion of the expression with a single variable (usually denoted as 'u') to make the integral more manageable.

This method is particularly useful when dealing with composite functions or expressions that involve chains of functions. The basic steps for u-substitution are as follows:

  1. Choose a suitable 'u' that simplifies the integral.
  2. Find du (the derivative of 'u') and express dx in terms of du.
  3. Rewrite the integral in terms of 'u.'
  4. Integrate with respect to 'u.'
  5. Substitute back in terms of 'x' if necessary.

👉 Trigonometric Functions

Trigonometric functions, like sine, cosine, and tangent, frequently appear in integrals. Understanding how to handle these functions is crucial. For example, when integrating trigonometric expressions, you may use trigonometric identities or specific trigonometric integrals.

Common trigonometric integrals include:

  • sin(x)dx=cos(x)+C∫\sin(x) dx = -\cos(x) + C
  • cos(x)dx=sin(x)+C∫\cos(x) dx = \sin(x) + C
  • tan(x)dx=lncos(x)+C∫\tan(x) dx = -\ln|\cos(x)| + C
  • cot(x)dx=lnsin(x)+C∫\cot(x) dx = \ln|\sin(x)| + C
  • sec(x)dx=lnsec(x)+tan(x)+C∫\sec(x) dx = \ln|\sec(x)+\tan(x)| + C
  • csc(x)dx=lncsc(x)+cot(x)+C∫\csc(x) dx = -\ln|\csc(x)+\cot(x)| + C

🙃 Inverse Trigonometric Functions

Antidifferentiation often involves inverse trigonometric functions like arcsin, arccos, and arctan. Knowing the antiderivatives of these functions and how to apply them is essential.

Here are some common inverse trigonometric integrals:

  • ddx(sin1(x))=11x2\frac {d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}
  • ddx(cos1(x))=11x2\frac {d}{dx}(\cos^{-1}(x)) = \frac{-1}{\sqrt{1-x^2}}
  • ddx(tan1(x))=11+x2\frac {d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2}
  • ddx(cot1(x))=11+x2\frac {d}{dx}(\cot^{-1}(x)) = \frac{-1}{1+x^2}
  • ddx(sec1(x))=1x1x2\frac {d}{dx}(\sec^{-1}(x)) = \frac{1}{x\sqrt{1-x^2}}
  • ddx(csc1(x))=1x1x2\frac {d}{dx}(\csc^{-1}(x)) = \frac{-1}{x\sqrt{1-x^2}}

🪴 Exponentials and Logarithms

Functions involving exponentials (exe^x) and natural logarithms (ln(x)\ln(x)) frequently appear in calculus problems. Familiarity with their antiderivatives is necessary for solving such integrals.

1.exdx=ex+C1. ∫e^x dx = e^x + C
2.1xdx=lnx+C2. ∫\frac{1}{x} dx = \ln| x | + C

Long Division

Long division can be useful when dealing with rational functions, where the degree of the numerator is equal to or higher than the degree of the denominator. It helps simplify the expression before antidifferentiation.

Here is the long division method for the expression:

(x2+2x3)/(x1) dx∫(x^2 + 2x - 3)/(x - 1) \ dx

🥇 Step 1: Perform Long Division

Perform long division to divide the numerator by the denominator:

x2+2x3x1=x3\frac{x^2+2x-3}{x-1} = x-3

🥈 Step 2: Write the Integral as a Sum Rewrite the integral as a sum of two integrals:

(x+3)dx(3(x1))dx∫(x + 3) dx - ∫(\frac{3}{(x - 1)}) dx

🥉 Step 3: Integrate Each Term Separately Now, integrate each term separately:

  • (x+3)dx=12x2+3x+C∫(x + 3) dx = \frac{1}{2}x^2 + 3x + C
  • (3(x1))dx=3lnx1+C∫(\frac{3}{(x - 1)}) dx = 3ln| x - 1 | + C

Step 4: Combine the Results Combine the results to find the antiderivative of the original expression:

(12)x2+3x3lnx1+C(\frac{1}{2})x^2 + 3x - 3ln| x - 1 | + C

🧊 Completing the Square

Completing the square is a technique often used with quadratic functions. It's crucial when you encounter expressions that can be transformed into a perfect square trinomial for easier integration.

Here's a step-by-step example of integrating a quadratic expression by completing the square, for…

(x2+4x+4)dx∫(x^2 + 4x + 4) dx

🥇 Step 1: Identify the Perfect Square Trinomial Recognize that the expression can be factored as a perfect square trinomial: (x+2)2(x + 2)^2.

🥈 Step 2: Rewrite the Integral Rewrite the integral using the perfect square trinomial:

(x2+4x+4)dx=(x+2)2dx∫(x^2 + 4x + 4) dx = ∫(x + 2)^2 dx

🥉 Step 3: Use the Power Rule for Antiderivatives Apply the power rule for antiderivatives to find the integral:

(x+2)2dx=(x+2)33+C∫(x + 2)^2 dx = \frac{(x + 2)^3}{3} + C

Step 4: Simplify Simplify the expression, resulting in its antiderivative:

(x+2)33+C\frac{(x + 2)^3}{3} + C

✏️ Integrating Practice Question

Let’s put those newly learned skills to good use! Find the antiderivative of the following expression:

(ex+2cos(x))dx∫(e^x + 2\cos(x)) dx

Be sure to try the question out yourself before moving onto the solution below!

✅ Solution to Practice Problem

When approaching this problem it is best to first recognize the terms in the expression and determine which technique to use for each term.

  • Term 1: exe^x (Antiderivative of exe^x is exe^x)
  • Term 2: 2cos(x)2\cos(x) (Antiderivative of cos(x)\cos(x) is sin(x)\sin(x))

Now, take the integral of each term individually.

  • Since the integral of exe^x is itself, term 1 is easy! exe^x
  • For term 2, the integral is 2sin(x)2\sin(x).

You got the two separate integrals! All that is left is to combine the results and add the +C+C term to the end.

(ex+2cos(x)) dx=ex+2sin(x)+C∫(e^x+2\cos(x))\ dx = e^x +2\sin(x)+C

So, the antiderivative of the given expression is ex+2sin(x)+Ce^x +2\sin(x)+C. Great work! 👏


⭐ Closing

Guess what?! You made it to the end of unit 6 in AP Calculus! Ready to move on to unit seven? It’s all about differential equations. Check it out here!

image.gif

Image Courtesy of Giphy