Evaluate the indefinite integral. ∫ 3 t t 2 + 7 d t \int_{}^{}3t\sqrt{t^2+7}dt

( t 2 + 7 ) 3 2 + C \left(t^2+7\right)^{\frac32}+C

( t 2 + 7 ) 2 3 + C \left(t^2+7\right)^{\frac23}+C

Evaluate the indefinite integral. ∫ 1 ( 3 x + 2 ) 5 d x \int_{}^{}\frac{1}{\left(3x+2\right)^5}dx

− 1 12 ( 3 x + 2 ) 4 + C -\frac{1}{12\left(3x+2\right)^4}+C

− 1 5 ( 3 x + 2 ) 4 + C -\frac{1}{5\left(3x+2\right)^4}+C

1 12 ( 3 x + 2 ) 4 + C \frac{1}{12\left(3x+2\right)^4}+C

1 5 ( 3 x + 2 ) 4 + C \frac{1}{5\left(3x+2\right)^4}+C

Evaluate the indefinite integral. ∫ θ ∙ s e c 2 ( 5 θ 2 + 1 ) d θ \int_{}^{}\theta\bullet sec^2\left(5\theta^2+1\right)d\theta

1 10 tan ⁡ ( 5 θ 2 + 1 ) + C \frac{1}{10}\tan\left(5\theta^2+1\right)+C

10 tan ⁡ ( 5 θ 2 + 1 ) + C 10\tan\left(5\theta^2+1\right)+C

10 s e c 2 ( 5 θ + 1 ) + C 10sec^2\left(5\theta+1\right)+C

1 20 θ 2 tan ⁡ ( 5 θ 2 + 1 ) + C \frac{1}{20}\theta^2\tan\left(5\theta^2+1\right)+C

Evaluate the indefinite integral. ∫ x ( 5 + x ) 79 d x \int_{}^{}x\left(5+x\right)^{79}dx

1 81 ( 5 + x ) 81 − 1 16 ( 5 + x ) 80 + C \frac{1}{81}\left(5+x\right)^{81}-\frac{1}{16}\left(5+x\right)^{80}+C

1 81 x 81 − 1 16 x 80 + C \frac{1}{81}x_{}^{81}-\frac{1}{16}x^{80}+C

1 81 ( 5 + x ) 81 − 1 80 ( 5 + x ) 80 + C \frac{1}{81}\left(5+x\right)^{81}-\frac{1}{80}\left(5+x\right)^{80}+C

1 81 x 81 − 1 80 x 80 + C \frac{1}{81}x^{81}-\frac{1}{80}x^{80}+C

Evaluate the indefinite integral. ∫ t t − 2 d t \int_{}^{}\frac{t}{\sqrt{t-2}}dt

2 3 ( t − 2 ) 3 2 + 4 ( t − 2 ) 1 2 + C \frac23\left(t-2\right)^{\frac32}+4\left(t-2\right)^{\frac12}+C

2 3 t 3 2 + 4 t 1 2 + C \frac23t^{\frac32}+4t^{\frac12}+C

2 3 ( t − 2 ) 2 3 + 2 + C \frac23\left(t-2\right)^{\frac23}+2+C

2 3 ( t + 2 ) 3 2 + 4 ( t + 2 ) 1 2 + C \frac23\left(t+2\right)^{\frac32}+4\left(t+2\right)^{\frac12}+C

Evaluate the indefinite integral. ∫ x ( x − 6 ) 5 d x \int_{}^{}\frac{x}{\left(x-6\right)^5}dx

− 1 3 ( x − 6 ) 3 − 3 2 ( x − 6 ) 4 + C -\frac{1}{3\left(x-6\right)^3}-\frac{3}{2\left(x-6\right)^4}+C

− x 6 ( x − 6 ) 6 − 3 x 6 ( x − 6 ) 6 + C -\frac{x}{6\left(x-6\right)^6}-\frac{3x}{6\left(x-6\right)^6}+C

− 1 3 ( x − 6 ) 6 − 3 2 ( x − 6 ) 6 + C -\frac{1}{3\left(x-6\right)^6}-\frac{3}{2\left(x-6\right)^6}+C

1 3 ( x − 6 ) 3 + 3 2 ( x − 6 ) 4 + C \frac{1}{3\left(x-6\right)^3}+\frac{3}{2\left(x-6\right)^4}+C

Evaluate the indefinite integral. ∫ s i n θ ( 1 + sin ⁡ θ ) 84 cos ⁡ θ d θ \int^{}sin\theta\left(1+\sin\theta\right)^{84}\cos\theta d\theta

1 86 ( 1 + sin ⁡ θ ) 86 − 1 85 ( 1 + sin ⁡ θ ) 85 + C \frac{1}{86}\left(1+\sin\theta\right)^{86}-\frac{1}{85}\left(1+\sin\theta\right)^{85}+C

− 1 85 ( sin ⁡ θ ) 85 − 1 86 ( sin ⁡ θ ) 86 -\frac{1}{85}\left(\sin\theta\right)^{85}-\frac{1}{86}\left(\sin\theta\right)^{86}

1 86 ( cos ⁡ θ ) 86 − 1 85 ( cos ⁡ θ ) 85 + C \frac{1}{86}\left(\cos\theta\right)^{86}-\frac{1}{85}\left(\cos\theta\right)^{85}+C

1 86 ( 1 + sin ⁡ θ ) 86 − 1 86 ( 1 + sin ⁡ θ ) 86 + C \frac{1}{86}\left(1+\sin\theta\right)^{86}-\frac{1}{86}\left(1+\sin\theta\right)^{86}+C

8. Definite Integrals

Substitution

8. Definite Integrals

Substitution: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

To evaluate integrals involving composite functions, the substitution method is essential. By choosing an appropriate substitution, such as $u = x + 3$ , and finding $du = dx$ , we can simplify the integral. For definite integrals, bounds can be transformed using the substitution, allowing for straightforward evaluation. This method enhances understanding of integrals and their applications in calculus, reinforcing concepts like the fundamental theorem of calculus and the power rule.

concept

Indefinite Integrals

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Indefinite Integrals Video Summary

In calculus, evaluating integrals involving composite functions can be simplified using a technique known as substitution, or u-substitution. This method is particularly useful when dealing with integrals that contain a function within another function, as well as a derivative of that inner function. The integral of the form $x^{2} + 1$ cubed times $2 x$ dx can be approached more easily by recognizing the structure of the integral.

To apply substitution, we first identify a suitable substitution for $u$ . In this case, we can let $u = x^{2} + 1$ . The next step is to find $du$ , which is the derivative of $u$ with respect to $x$ multiplied by $dx$ . Calculating this gives us $du = 2 x dx$ , which matches the term in our integral.

Substituting $u$ into the integral transforms it into a simpler form: $u^{3} du$ . This integral can now be evaluated using the power rule, resulting in $1 / 4 u^{4} + C$ . Finally, substituting back for $u$ gives us the final answer: $1 / 4 (^{x^{2} + 1)} 4 + C$ .

In another example, consider the integral of $4 \sqrt{4 x - 1} dx$ . Here, we again choose $u = 4 x - 1$ , leading to $du = 4 dx$ . To adjust for the constant, we multiply the integral by $1 / 4$ to maintain its value, resulting in $1 / 4 \sqrt{u} du$ . Evaluating this integral using the power rule yields $1 / 6 u^{3} + C$ . Substituting back for $u$ gives the final result: $1 / 6 (^{4 x - 1)} 3 / 2 + C$ .

In summary, the substitution method streamlines the process of evaluating integrals involving composite functions. By carefully selecting $u$ and determining $du$ , the integral can be transformed into a more manageable form, allowing for straightforward integration and eventual substitution back to the original variable.

example

Indefinite Integrals Example 1

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Indefinite Integrals Example 1 Video Summary

To evaluate the integral of the expression $ (2x^3 + x + 7)^5 (6x^2 + 1) \, dx $ using substitution, we start by identifying a suitable substitution. We choose $ u = 2x^3 + x + 7 $, which is the inner function raised to a power. Next, we compute the derivative $ du $ by differentiating $ u $ with respect to $ x $. The derivative is given by:

\[du = (6x^2 + 1) \, dx\]

This allows us to rewrite the integral in terms of $ u $ and $ du $. Substituting, we have:

\[\int (2x^3 + x + 7)^5 (6x^2 + 1) \, dx = \int u^5 \, du\]

Now, we can integrate $ u^5 $ using the power rule. The integral becomes:

\[\int u^5 \, du = \frac{1}{6} u^6 + C\]

Substituting back for $ u $, we find:

\[\frac{1}{6} (2x^3 + x + 7)^6 + C\]

This is our final result for the integral. To verify our solution, we differentiate the result:

\[\frac{d}{dx} \left( \frac{1}{6} (2x^3 + x + 7)^6 + C \right)\]

Using the chain rule, we differentiate the outer function first, which gives:

\[\frac{1}{6} \cdot 6 (2x^3 + x + 7)^5 \cdot \frac{d}{dx}(2x^3 + x + 7)\]

The derivative of $ 2x^3 + x + 7 $ is $ 6x^2 + 1 $. Thus, we have:

\[(2x^3 + x + 7)^5 (6x^2 + 1)\]

This matches the original integrand, confirming that our integration was performed correctly. Therefore, the evaluated integral is:

\[\frac{1}{6} (2x^3 + x + 7)^6 + C\]

example

Indefinite Integrals Example 2

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Indefinite Integrals Example 2 Video Summary

To evaluate the indefinite integral of the cosine of $ t $ multiplied by the sine of $ t $ raised to the ninety-ninth power, we can utilize substitution to simplify the process. The integral we are working with is:

\[\int \cos(t) \sin^{99}(t) \, dt\]

First, we choose a substitution. A good candidate for $ u $ is the sine function, since it is raised to a power. Let:

\[u = \sin(t)\]

Then, we find the differential $ du $. The derivative of $ \sin(t) $ is $ \cos(t) $, so:

\[du = \cos(t) \, dt\]

Now, we can rewrite the integral in terms of $ u $ and $ du $. Rearranging the original integral gives us:

\[\int \sin^{99}(t) \cos(t) \, dt\]

Substituting $ u $ for $ \sin(t) $ and $ du $ for $ \cos(t) \, dt $, we have:

\[\int u^{99} \, du\]

Next, we integrate with respect to $ u $ using the power rule. The power rule states that:

\[\int u^n \, du = \frac{u^{n+1}}{n+1} + C\]

Applying this to our integral:

\[\int u^{99} \, du = \frac{u^{100}}{100} + C\]

Now, we substitute back $ u = \sin(t) $ into our result:

\[\frac{\sin^{100}(t)}{100} + C\]

To present this in a more conventional format, we can rewrite it as:

\[\frac{1}{100} \sin^{100}(t) + C\]

This is the final answer for the indefinite integral:

\[\int \cos(t) \sin^{99}(t) \, dt = \frac{1}{100} \sin^{100}(t) + C\]

Problem

Evaluate the indefinite integral.
$\int_{}^{} 3 t \sqrt{t^{2} + 7} d t$

$t^{\frac{3}{2}} + C$

${(t^{2} + 7)}^{\frac{2}{3}} + C$

${(t^{2} + 7)}^{\frac{3}{2}} + C$

$t^{\frac{2}{3}} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} \frac{1}{{(3 x + 2)}^{5}} d x$

$- \frac{1}{12 {(3 x + 2)}^{4}} + C$

$- \frac{1}{5 {(3 x + 2)}^{4}} + C$

$\frac{1}{12 {(3 x + 2)}^{4}} + C$

$\frac{1}{5 {(3 x + 2)}^{4}} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} θ ∙ s e c^{2} (5 θ^{2} + 1) d θ$

$10 \tan (5 θ^{2} + 1) + C$

$\frac{1}{10} \tan (5 θ^{2} + 1) + C$

$10 s e c^{2} (5 θ + 1) + C$

$\frac{1}{20} θ^{2} \tan (5 θ^{2} + 1) + C$

concept

Substitution With an Extra Variable

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Substitution With an Extra Variable Video Summary

In the process of evaluating integrals using substitution, it's essential to recognize that sometimes the integral may contain additional factors that need to be addressed. This summary will guide you through the steps of performing a substitution effectively, ensuring that all variables are accounted for.

To illustrate this, consider the integral of $ x \sqrt{x + 3} \, dx $. The first step is to choose a substitution for $ u $. A good choice here is $ u = x + 3 $, which leads to $ du = dx $ since the derivative of $ x + 3 $ is 1. This substitution simplifies the integral, but we still have the variable $ x $ present.

To eliminate $ x $ from the integral, we can express $ x $ in terms of $ u $. Rearranging the substitution gives us $ x = u - 3 $. Substituting this back into the integral transforms it into:

\[\int (u - 3) \sqrt{u} \, du\]

Next, we simplify the integral. The square root of $ u $ can be rewritten as $ u^{1/2} $. Distributing $ (u - 3) $ gives us:

\[\int (u^{1} \cdot u^{1/2} - 3u^{1/2}) \, du = \int (u^{3/2} - 3u^{1/2}) \, du\]

Now, we can apply the power rule for integration. For $ u^{3/2} $, we add 1 to the exponent, resulting in $ u^{5/2} $, and divide by $ 5/2 $, which is equivalent to multiplying by $ \frac{2}{5} $. Thus, the integral becomes:

\[\frac{2}{5} u^{5/2}\]

For the term $ -3u^{1/2} $, we again apply the power rule, yielding:

\[-3 \cdot \frac{2}{3} u^{3/2} = -2u^{3/2}\]

Combining these results, we have:

\[\frac{2}{5} u^{5/2} - 2u^{3/2} + C\]

Finally, we substitute back $ u = x + 3 $ to express the integral in terms of $ x $. This gives us the final answer:

\[\frac{2}{5} (x + 3)^{5/2} - 2(x + 3)^{3/2} + C\]

This process highlights the importance of carefully managing all variables during substitution, ensuring that the integral is fully expressed in terms of the new variable before integrating. Mastering this technique will enhance your ability to tackle a wide range of integrals effectively.

Problem

Evaluate the indefinite integral.
$\int_{}^{} x {(5 + x)}^{79} d x$

$\frac{1}{81} {(5 + x)}^{81} - \frac{1}{16} {(5 + x)}^{80} + C$

$\frac{1}{81} x_{}^{81} - \frac{1}{16} x^{80} + C$

$\frac{1}{81} {(5 + x)}^{81} - \frac{1}{80} {(5 + x)}^{80} + C$

$\frac{1}{81} x^{81} - \frac{1}{80} x^{80} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} \frac{t}{\sqrt{t - 2}} d t$

$\frac{2}{3} t^{\frac{3}{2}} + 4 t^{\frac{1}{2}} + C$

$\frac{2}{3} {(t - 2)}^{\frac{2}{3}} + 2 + C$

$\frac{2}{3} {(t + 2)}^{\frac{3}{2}} + 4 {(t + 2)}^{\frac{1}{2}} + C$

$\frac{2}{3} {(t - 2)}^{\frac{3}{2}} + 4 {(t - 2)}^{\frac{1}{2}} + C$

Problem

Evaluate the indefinite integral.
$\int_{}^{} \frac{x}{{(x - 6)}^{5}} d x$

$- \frac{x}{6 {(x - 6)}^{6}} - \frac{3 x}{6 {(x - 6)}^{6}} + C$

$- \frac{1}{3 {(x - 6)}^{6}} - \frac{3}{2 {(x - 6)}^{6}} + C$

$- \frac{1}{3 {(x - 6)}^{3}} - \frac{3}{2 {(x - 6)}^{4}} + C$

$\frac{1}{3 {(x - 6)}^{3}} + \frac{3}{2 {(x - 6)}^{4}} + C$

Problem

Evaluate the indefinite integral.
$\int^{} s i n θ {(1 + \sin θ)}^{84} \cos θ d θ$

$- \frac{1}{85} {(\sin θ)}^{85} - \frac{1}{86} {(\sin θ)}^{86}$

$\frac{1}{86} {(\cos θ)}^{86} - \frac{1}{85} {(\cos θ)}^{85} + C$

$\frac{1}{86} {(1 + \sin θ)}^{86} - \frac{1}{85} {(1 + \sin θ)}^{85} + C$

$\frac{1}{86} {(1 + \sin θ)}^{86} - \frac{1}{86} {(1 + \sin θ)}^{86} + C$

concept

Definite Integrals

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Definite Integrals Video Summary

In evaluating definite integrals using substitution, we can apply similar techniques as with indefinite integrals, with the key difference being the presence of bounds. There are two effective methods to handle these bounds during the evaluation process.

Consider the definite integral from 0 to 2 of $ (x^2 + 1)^3 \cdot 2x \, dx $. The first method involves treating the integral as if it were indefinite. We start by selecting $ u = x^2 + 1 $, which gives us $ du = 2x \, dx $. This allows us to rewrite the integral as $ \int u^3 \, du $. We then find the antiderivative, which is $ \frac{1}{4} u^4 + C $. Substituting back for $ u $, we have $ \frac{1}{4} (x^2 + 1)^4 + C $. Since this is a definite integral, we evaluate the antiderivative at the bounds 0 and 2, leading to the expression $ \frac{1}{4} \left(2^2 + 1\right)^4 - \frac{1}{4} \left(0^2 + 1\right)^4 $. Calculating this gives a final result of 156.

The second method involves transforming the bounds along with the integral. Using the same integral, we again set $ u = x^2 + 1 $ and $ du = 2x \, dx $, rewriting the integral as $ \int u^3 \, du $. However, this time we also change the bounds: for the lower bound, when $ x = 0 $, $ u = 0^2 + 1 = 1 $; for the upper bound, when $ x = 2 $, $ u = 2^2 + 1 = 5 $. Thus, the new bounds for the integral are from 1 to 5. We then integrate $ \int u^3 \, du $ to get $ \frac{1}{4} u^4 $ and evaluate it at the new bounds, resulting in $ \frac{1}{4} (5^4 - 1^4) $. This also yields a final answer of 156.

Both methods are valid and yield the same result. The choice of method depends on personal preference, but it is crucial to understand how to properly apply the bounds in each case. The first method treats the integral as indefinite before applying the bounds, while the second method incorporates the bounds into the substitution process from the start. Mastery of these techniques will enhance your ability to tackle a variety of definite integrals using substitution.

Problem

Evaluate the definite integral.
$\int_{0}^{1} \frac{t}{\sqrt{t^{2} + 1}} d t$

1.414

0.414

0.707

Problem

Evaluate the definite integral.
$\int_{1}^{2} (x - 3) {(x^{2} - 6 x)}^{2} d x$

-64.5

533

-2.33

-1.17

Problem

Evaluate the definite integral.
$\int_{- \frac{π}{4}}^{\frac{π}{4}} \tan y$ $s e c^{2} y d y$

0.308

$π$

$pi over 4$

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Here’s what students ask on this topic:

What is the substitution method in calculus, and how is it used to evaluate integrals?

The substitution method in calculus is a technique used to simplify the evaluation of integrals, especially those involving composite functions. It involves choosing a substitution, typically denoted as $u$ , to replace a part of the integrand, and then finding $d u$ to replace $d x$ . This transforms the integral into a simpler form, often allowing the use of basic integration rules like the power rule. For definite integrals, the bounds are also transformed using the substitution, facilitating straightforward evaluation. This method is crucial for understanding integrals and their applications, reinforcing concepts such as the fundamental theorem of calculus.

How do you choose the correct substitution for an integral?

Choosing the correct substitution for an integral involves identifying the 'inside function' within the integrand. This function is often found inside parentheses, under a radical, or in an exponent. Once identified, it is set as $u$ . The derivative of this function, multiplied by $d x$ , gives $d u$ . The goal is to rewrite the integral entirely in terms of $u$ and $d u$ . If the substitution doesn't perfectly match the integrand, adjustments like multiplying by constants or rearranging equations may be necessary. This process simplifies the integral, making it easier to evaluate.

What are the steps to evaluate a definite integral using substitution?

To evaluate a definite integral using substitution, follow these steps: First, choose a substitution $u$ and find $d u$ . Next, rewrite the integral in terms of $u$ and $d u$ , transforming the bounds by substituting the original limits into the function of $u$ . Then, integrate with respect to $u$ . Finally, evaluate the antiderivative at the transformed bounds using the fundamental theorem of calculus. This method allows for a straightforward evaluation of definite integrals, ensuring accuracy and efficiency.

How does the substitution method relate to the chain rule in calculus?

The substitution method is essentially the reverse process of the chain rule in calculus. While the chain rule is used to differentiate composite functions, substitution is used to integrate them. In differentiation, the chain rule involves taking the derivative of the outer function and multiplying it by the derivative of the inner function. In integration, substitution involves identifying the inner function and its derivative, then transforming the integral into a simpler form. This relationship highlights the interconnectedness of differentiation and integration, reinforcing the concept that integration is the reverse operation of differentiation.

What are common mistakes to avoid when using substitution in integrals?

Common mistakes when using substitution in integrals include choosing an incorrect substitution, failing to transform the bounds in definite integrals, and neglecting to replace the original variable after integration. It's crucial to identify the correct 'inside function' for substitution and ensure that $d u$ matches the integrand. For definite integrals, bounds must be transformed using the substitution. After integrating, always replace $u$ with the original variable to express the final answer in terms of the original variable. Attention to these details ensures accurate and efficient evaluation of integrals.

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Substitution: Videos & Practice Problems

Indefinite Integrals

Indefinite Integrals Video Summary

Indefinite Integrals Example 1

Indefinite Integrals Example 1 Video Summary

Indefinite Integrals Example 2

Indefinite Integrals Example 2 Video Summary

Evaluate the indefinite integral.∫3tt2+7dt\int_{}^{}3t\sqrt{t^2+7}dt

Evaluate the indefinite integral.∫1(3x+2)5dx\int_{}^{}\frac{1}{\left(3x+2\right)^5}dx

Evaluate the indefinite integral.∫θ∙sec2(5θ2+1)dθ\int_{}^{}\theta\bullet sec^2\left(5\theta^2+1\right)d\theta

Substitution With an Extra Variable

Substitution With an Extra Variable Video Summary

Evaluate the indefinite integral.∫x(5+x)79dx\int_{}^{}x\left(5+x\right)^{79}dx

Evaluate the indefinite integral.∫tt−2dt\int_{}^{}\frac{t}{\sqrt{t-2}}dt

Evaluate the indefinite integral.∫x(x−6)5dx\int_{}^{}\frac{x}{\left(x-6\right)^5}dx

Evaluate the indefinite integral.∫sinθ(1+sinθ)84cosθdθ\int^{}sin\theta\left(1+\sin\theta\right)^{84}\cos\theta d\theta

Definite Integrals

Definite Integrals Video Summary

Evaluate the definite integral.∫01tt2+1dt\int_0^1\frac{t}{\sqrt{t^2+1}}dt

Evaluate the definite integral.∫12(x−3)(x2−6x)2dx\int_1^2\left(x-3\right)\left(x^2-6x\right)^2dx

Evaluate the definite integral.∫−π4π4tany\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan y sec2ydysec^2ydy

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Here’s what students ask on this topic:

What is the substitution method in calculus, and how is it used to evaluate integrals?

How do you choose the correct substitution for an integral?

What are the steps to evaluate a definite integral using substitution?

How does the substitution method relate to the chain rule in calculus?

What are common mistakes to avoid when using substitution in integrals?

Your Calculus tutors

Evaluate the indefinite integral.
$\int_{}^{} 3 t \sqrt{t^{2} + 7} d t$

Evaluate the indefinite integral.
$\int_{}^{} \frac{1}{{(3 x + 2)}^{5}} d x$

Evaluate the indefinite integral.
$\int_{}^{} θ ∙ s e c^{2} (5 θ^{2} + 1) d θ$

Evaluate the indefinite integral.
$\int_{}^{} x {(5 + x)}^{79} d x$

Evaluate the indefinite integral.
$\int_{}^{} \frac{t}{\sqrt{t - 2}} d t$

Evaluate the indefinite integral.
$\int_{}^{} \frac{x}{{(x - 6)}^{5}} d x$

Evaluate the indefinite integral.
$\int^{} s i n θ {(1 + \sin θ)}^{84} \cos θ d θ$

Evaluate the definite integral.
$\int_{0}^{1} \frac{t}{\sqrt{t^{2} + 1}} d t$

Evaluate the definite integral.
$\int_{1}^{2} (x - 3) {(x^{2} - 6 x)}^{2} d x$

Evaluate the definite integral.
$\int_{- \frac{π}{4}}^{\frac{π}{4}} \tan y$ $s e c^{2} y d y$