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Учебные пособия > Calculus Volume 1

Substitution with Indefinite Integrals

Learning Objectives

  • Use substitution to evaluate indefinite integrals.
  • Integrate functions involving exponential functions.
  • Integrate functions involving logarithmic functions.
  • Integrate functions resulting in inverse trigonometric functions.

The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.

At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form [latex]f\left[g(x)\right]{g}^{\prime }(x)dx.[/latex] For example, in the integral [latex]\int {({x}^{2}-3)}^{3}2xdx,[/latex] we have [latex]f(x)={x}^{3},g(x)={x}^{2}-3,[/latex] and [latex]g\text{‘}(x)=2x.[/latex] Then,

[latex]f\left[g(x)\right]{g}^{\prime }(x)={({x}^{2}-3)}^{3}(2x),[/latex]

and we see that our integrand is in the correct form.

The method is called substitution because we substitute part of the integrand with the variable [latex]u[/latex] and part of the integrand with du. It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.

Substitution with Indefinite Integrals

Let [latex]u=g(x),,[/latex] where [latex]{g}^{\prime }(x)[/latex] is continuous over an interval, let [latex]f(x)[/latex] be continuous over the corresponding range of [latex]g[/latex], and let [latex]F(x)[/latex] be an antiderivative of [latex]f(x).[/latex] Then,

[latex]\begin{array}{cc}\int f\left[g(x)\right]{g}^{\prime }(x)dx\hfill & =\int f(u)du\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end{array}[/latex]

Proof

Let [latex]f[/latex], [latex]g[/latex], [latex]u[/latex], and F be as specified in the theorem. Then

[latex]\begin{array}{cc}\frac{d}{dx}F(g(x))\hfill & ={F}^{\prime }(g(x)){g}^{\prime }(x)\hfill \\ & =f\left[g(x)\right]{g}^{\prime }(x).\hfill \end{array}[/latex]

Integrating both sides with respect to [latex]x[/latex], we see that

[latex]\int f\left[g(x)\right]{g}^{\prime }(x)dx=F(g(x))+C.[/latex]

If we now substitute [latex]u=g(x),[/latex] and [latex]du=g\text{‘}(x)dx,[/latex] we get

[latex]\begin{array}{cc}\int f\left[g(x)\right]{g}^{\prime }(x)dx\hfill & =\int f(u)du\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end{array}[/latex]

□

Returning to the problem we looked at originally, we let [latex]u={x}^{2}-3[/latex] and then [latex]du=2xdx.[/latex] Rewrite the integral in terms of [latex]u[/latex]:

[latex]{\int \underset{u}{\underbrace{({x}^{2}-3)}}}^{3}\underset{du}{\underbrace{(2xdx)}}=\int {u}^{3}du.[/latex]
Using the power rule for integrals, we have
[latex]\int {u}^{3}du=\frac{{u}^{4}}{4}+C.[/latex]

Substitute the original expression for [latex]x[/latex] back into the solution:

[latex]\frac{{u}^{4}}{4}+C=\frac{{({x}^{2}-3)}^{4}}{4}+C.[/latex]

We can generalize the procedure in the following Problem-Solving Strategy.

Problem-Solving Strategy: Integration by Substitution

  1. Look carefully at the integrand and select an expression [latex]g(x)[/latex] within the integrand to set equal to [latex]u[/latex]. Let’s select [latex]g(x).[/latex] such that [latex]{g}^{\prime }(x)[/latex] is also part of the integrand.
  2. Substitute [latex]u=g(x)[/latex] and [latex]du={g}^{\prime }(x)dx.[/latex] into the integral.
  3. We should now be able to evaluate the integral with respect to [latex]u[/latex]. If the integral can’t be evaluated we need to go back and select a different expression to use as [latex]u[/latex].
  4. Evaluate the integral in terms of [latex]u[/latex].
  5. Write the result in terms of [latex]x[/latex] and the expression [latex]g(x).[/latex]

Evaluating an inDefinite Integral Using Substitution

Use substitution to find the antiderivative of [latex]\int 6x{(3{x}^{2}+4)}^{4}dx.[/latex]

Answer:

The first step is to choose an expression for [latex]u[/latex]. We choose [latex]u=3{x}^{2}+4.[/latex] because then [latex]du=6xdx.,[/latex] and we already have du in the integrand. Write the integral in terms of [latex]u[/latex]:

[latex]\int 6x{(3{x}^{2}+4)}^{4}dx=\int {u}^{4}du.[/latex]

Remember that du is the derivative of the expression chosen for [latex]u[/latex], regardless of what is inside the integrand. Now we can evaluate the integral with respect to [latex]u[/latex]:

[latex]\begin{array}{ll}\int {u}^{4}du\hfill & =\frac{{u}^{5}}{5}+C\hfill \\ \\ \\ & =\frac{{(3{x}^{2}+4)}^{5}}{5}+C.\hfill \end{array}[/latex]

Analysis

We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for C of 1, we let [latex]y=\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.[/latex] We have

[latex]y=\frac{1}{5}{(3{x}^{2}+4)}^{5}+1,[/latex]

so

[latex]\begin{array}{}\\ \hfill {y}^{\prime }& =(\frac{1}{5})5{(3{x}^{2}+4)}^{4}6x\hfill \\ & =6x{(3{x}^{2}+4)}^{4}.\hfill \end{array}[/latex]

This is exactly the expression we started with inside the integrand.

Use substitution to find the antiderivative of [latex]\int 3{x}^{2}{({x}^{3}-3)}^{2}dx.[/latex]

Answer:

[latex]\int 3{x}^{2}{({x}^{3}-3)}^{2}dx=\frac{1}{3}{({x}^{3}-3)}^{3}+C[/latex]

Hint

Let [latex]u={x}^{3}-3.[/latex]

Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.

Using Substitution with Alteration

Use substitution to find the antiderivative of [latex]\int z\sqrt{{z}^{2}-5}dz.[/latex]

Answer:

Rewrite the integral as [latex]\int z{({z}^{2}-5)}^{1\text{/}2}dz.[/latex] Let [latex]u={z}^{2}-5[/latex] and [latex]du=2zdz.[/latex] Now we have a problem because [latex]du=2zdz[/latex] and the original expression has only [latex]zdz.[/latex] We have to alter our expression for du or the integral in [latex]u[/latex] will be twice as large as it should be. If we multiply both sides of the du equation by [latex]\frac{1}{2}.[/latex] we can solve this problem. Thus,

[latex]\begin{array}{}\\ \hfill u& ={z}^{2}-5\hfill \\ \hfill du& =2zdz\hfill \\ \hfill \frac{1}{2}du& =\frac{1}{2}(2z)dz=zdz.\hfill \end{array}[/latex]

Write the integral in terms of [latex]u[/latex], but pull the [latex]\frac{1}{2}[/latex] outside the integration symbol:

[latex]\int z{({z}^{2}-5)}^{1\text{/}2}dz=\frac{1}{2}\int {u}^{1\text{/}2}du.[/latex]

Integrate the expression in [latex]u[/latex]:

[latex]\begin{array}{}\\ \frac{1}{2}\int {u}^{1\text{/}2}du\hfill & =(\frac{1}{2})\frac{{u}^{3\text{/}2}}{\frac{3}{2}}+C\hfill \\ \\ & =(\frac{1}{2})(\frac{2}{3}){u}^{3\text{/}2}+C\hfill \\ & =\frac{1}{3}{u}^{3\text{/}2}+C\hfill \\ & =\frac{1}{3}{({z}^{2}-5)}^{3\text{/}2}+C.\hfill \end{array}[/latex]   

Use substitution to find the antiderivative of [latex]\int {x}^{2}{({x}^{3}+5)}^{9}dx.[/latex]

Answer:

[latex]\frac{{({x}^{3}+5)}^{10}}{30}+C[/latex]

Hint

Multiply the du equation by [latex]\frac{1}{3}.[/latex] 

Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral [latex]\int \frac{ \sin t}{{ \cos }^{3}t}dt.[/latex]

Answer:

We know the derivative of [latex] \cos t[/latex] is [latex]\text{−} \sin t,[/latex] so we set [latex]u= \cos t.[/latex] Then [latex]du=\text{−} \sin tdt.[/latex] Substituting into the integral, we have

[latex]\int \frac{ \sin t}{{ \cos }^{3}t}dt=\text{−}\int \frac{du}{{u}^{3}}.[/latex]

Evaluating the integral, we get

[latex]\begin{array}{}\\ \\ \text{−}\int \frac{du}{{u}^{3}}\hfill & =\text{−}\int {u}^{-3}du\hfill \\ & =\text{−}(-\frac{1}{2}){u}^{-2}+C.\hfill \end{array}[/latex]

Putting the answer back in terms of [latex]t[/latex], we get

[latex]\begin{array}{cc}\int \frac{ \sin t}{{ \cos }^{3}t}dt\hfill & =\frac{1}{2{u}^{2}}+C\hfill \\ \\ & =\frac{1}{2{ \cos }^{2}t}+C.\hfill \end{array}[/latex]

Use substitution to evaluate the integral [latex]\int \frac{ \cos t}{{ \sin }^{2}t}dt.[/latex]

Answer:

[latex]-\frac{1}{ \sin t}+C[/latex]

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, [latex]u[/latex] should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of [latex]u[/latex]. This technique should become clear in the next example.

Finding an Antiderivative Using [latex]u[/latex]-Substitution

Use substitution to find the antiderivative of [latex]\int \frac{x}{\sqrt{x-1}}dx.[/latex]

Answer:

If we let [latex]u=x-1,[/latex] then [latex]du=dx.[/latex] But this does not account for the [latex]x[/latex] in the numerator of the integrand. We need to express [latex]x[/latex] in terms of [latex]u[/latex]. If [latex]u=x-1,[/latex] then [latex]x=u+1.[/latex] Now we can rewrite the integral in terms of [latex]u[/latex]:

[latex]\begin{array}{ll}\int \frac{x}{\sqrt{x-1}}dx\hfill & =\int \frac{u+1}{\sqrt{u}}du\hfill \\ \\ & =\int \sqrt{u}+\frac{1}{\sqrt{u}}du\hfill \\ & =\int ({u}^{1\text{/}2}+{u}^{-1\text{/}2})du.\hfill \end{array}[/latex]

Then we integrate in the usual way, replace [latex]u[/latex] with the original expression, and factor and simplify the result. Thus,

[latex]\begin{array}{cc}\int ({u}^{1\text{/}2}+{u}^{-1\text{/}2})du\hfill & =\frac{2}{3}{u}^{3\text{/}2}+2{u}^{1\text{/}2}+C\hfill \\ \\ & =\frac{2}{3}{(x-1)}^{3\text{/}2}+2{(x-1)}^{1\text{/}2}+C\hfill \\ & ={(x-1)}^{1\text{/}2}\left[\frac{2}{3}(x-1)+2\right]+C\hfill \\ & ={(x-1)}^{1\text{/}2}(\frac{2}{3}x-\frac{2}{3}+\frac{6}{3})\hfill \\ & ={(x-1)}^{1\text{/}2}(\frac{2}{3}x+\frac{4}{3})\hfill \\ & =\frac{2}{3}{(x-1)}^{1\text{/}2}(x+2)+C.\hfill \end{array}[/latex]

Use substitution to evaluate the indefinite integral [latex]\int { \cos }^{3}t \sin tdt.[/latex]

Answer:

[latex]-\frac{{ \cos }^{4}t}{4}+C[/latex]

Integrals of Exponential Functions

The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, [latex]y={e}^{x},[/latex] is its own derivative and its own integral.

Rule: Integrals of Exponential Functions

Exponential functions can be integrated using the following formulas.

[latex]\begin{array}{ccc}\int {e}^{x}dx\hfill & =\hfill & {e}^{x}+C\hfill \\ \int {a}^{x}dx\hfill & =\hfill & \frac{{a}^{x}}{\text{ln}a}+C\hfill \end{array}[/latex]

Finding an Antiderivative of an Exponential Function

Find the antiderivative of the exponential function [latex]e[/latex]−[latex]x[/latex].

Answer:

Use substitution, setting [latex]u=\text{−}x,[/latex] and then [latex]du=-1dx.[/latex] Multiply the du equation by −1, so you now have [latex]\text{−}du=dx.[/latex] Then,

[latex]\begin{array}{cc}\int {e}^{\text{−}x}dx\hfill & =\text{−}\int {e}^{u}du\hfill \\ \\ & =\text{−}{e}^{u}+C\hfill \\ & =\text{−}{e}^{\text{−}x}+C.\hfill \end{array}[/latex]

Find the antiderivative of the function using substitution: [latex]{x}^{2}{e}^{-2{x}^{3}}.[/latex]

Answer:

[latex]\int {x}^{2}{e}^{-2{x}^{3}}dx=-\frac{1}{6}{e}^{-2{x}^{3}}+C[/latex]

Hint

Let [latex]u[/latex] equal the exponent on [latex]e[/latex].

A common mistake when dealing with exponential expressions is treating the exponent on [latex]e[/latex] the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent on [latex]e[/latex]. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we’re using the right rules for the functions we’re integrating.

Square Root of an Exponential Function

Find the antiderivative of the exponential function [latex]{e}^{x}\sqrt{1+{e}^{x}}.[/latex]

Answer:

First rewrite the problem using a rational exponent:

[latex]\int {e}^{x}\sqrt{1+{e}^{x}}dx=\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx.[/latex]
Using substitution, choose [latex]u=1+{e}^{x}.u=1+{e}^{x}.[/latex] Then, [latex]du={e}^{x}dx.[/latex] We have ((Figure))
[latex]\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx=\int {u}^{1\text{/}2}du.[/latex]

Then

[latex]\int {u}^{1\text{/}2}du=\frac{{u}^{3\text{/}2}}{3\text{/}2}+C=\frac{2}{3}{u}^{3\text{/}2}+C=\frac{2}{3}{(1+{e}^{x})}^{3\text{/}2}+C.[/latex]
A graph of the function f(x) = e^x * sqrt(1 + e^x), which is an increasing concave up curve, over [-3, 1]. It begins close to the x axis in quadrant two, crosses the y axis at (0, sqrt(2)), and continues to increase rapidly. Figure 1. The graph shows an exponential function times the square root of an exponential function.

Find the antiderivative of [latex]{e}^{x}{(3{e}^{x}-2)}^{2}.[/latex]

[latex]\int {e}^{x}{(3{e}^{x}-2)}^{2}dx=\frac{1}{9}{(3{e}^{x}-2)}^{3}[/latex]

Hint

Let [latex]u=3{e}^{x}-2u=3{e}^{x}-2.[/latex]

Using Substitution with an Exponential Function

Use substitution to evaluate the indefinite integral [latex]\int 3{x}^{2}{e}^{2{x}^{3}}dx.[/latex]

Answer:

Here we choose to let [latex]u[/latex] equal the expression in the exponent on [latex]e[/latex]. Let [latex]u=2{x}^{3}[/latex] and [latex]du=6{x}^{2}dx..[/latex] Again, du is off by a constant multiplier; the original function contains a factor of 3[latex]x[/latex]2, not 6[latex]x[/latex]2. Multiply both sides of the equation by [latex]\frac{1}{2}[/latex] so that the integrand in [latex]u[/latex] equals the integrand in [latex]x[/latex]. Thus,

[latex]\int 3{x}^{2}{e}^{2{x}^{3}}dx=\frac{1}{2}\int {e}^{u}du.[/latex]

Integrate the expression in [latex]u[/latex] and then substitute the original expression in [latex]x[/latex] back into the [latex]u[/latex] integral:

[latex]\frac{1}{2}\int {e}^{u}du=\frac{1}{2}{e}^{u}+C=\frac{1}{2}{e}^{2{x}^{3}}+C.[/latex]

Evaluate the indefinite integral [latex]\int 2{x}^{3}{e}^{{x}^{4}}dx.[/latex]

Answer:

[latex]\int 2{x}^{3}{e}^{{x}^{4}}dx=\frac{1}{2}{e}^{{x}^{4}}[/latex]

Hint

Let [latex]u={x}^{4}.[/latex]

Integrals Involving Logarithmic Functions

Integrating functions of the form [latex]f(x)={x}^{-1}[/latex] result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as [latex]f(x)=\text{ln}x[/latex] and [latex]f(x)={\text{log}}_{a}x,[/latex] are also included in the rule.

Rule: Integration Formulas Involving Logarithmic Functions

The following formulas can be used to evaluate integrals involving logarithmic functions.

[latex]\begin{array}{ccc}\hfill \int {x}^{-1}dx& =\hfill & \text{ln}|x|+C\hfill \\ \hfill \int \text{ln}xdx& =\hfill & x\text{ln}x-x+C=x(\text{ln}x-1)+C\hfill \\ \hfill \int {\text{log}}_{a}xdx& =\hfill & \frac{x}{\text{ln}a}(\text{ln}x-1)+C\hfill \end{array}[/latex]

Finding an Antiderivative Involving [latex]\text{ln}x[/latex]

Find the antiderivative of the function [latex]\frac{3}{x-10}.[/latex]

Answer:

First factor the 3 outside the integral symbol. Then use the [latex]u[/latex]−1 rule. Thus,

[latex]\begin{array}{ll}\int \frac{3}{x-10}dx\hfill & =3\int \frac{1}{x-10}dx\hfill \\ \\ \\ & =3\int \frac{du}{u}\hfill \\ & =3\text{ln}|u|+C\hfill \\ & =3\text{ln}|x-10|+C,x\ne 10.\hfill \end{array}[/latex]

See (Figure).

A graph of the function f(x) = 3 / (x – 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity. Figure 3. The domain of this function is [latex]x\ne 10.[/latex]

Find the antiderivative of [latex]\frac{1}{x+2}.[/latex]

Answer:

[latex]\text{ln}|x+2|+C[/latex]

Hint

Follow the pattern from (Figure) to solve the problem.

Finding an Antiderivative of a Rational Function

Find the antiderivative of [latex]\frac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.[/latex]

This can be rewritten as [latex]\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.[/latex] Use substitution. Let [latex]u={x}^{4}+3{x}^{2},[/latex] then [latex]du=4{x}^{3}+6x.[/latex] Alter du by factoring out the 2. Thus,
[latex]\begin{array}{}\\ \hfill du& =\hfill & (4{x}^{3}+6x)dx\hfill \\ & =\hfill & 2(2{x}^{3}+3x)dx\hfill \\ \hfill \frac{1}{2}du& =\hfill & (2{x}^{3}+3x)dx.\hfill \end{array}[/latex]

Rewrite the integrand in [latex]u[/latex]:

[latex]\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx=\frac{1}{2}\int {u}^{-1}du.[/latex]

Then we have

[latex]\begin{array}{ll}\frac{1}{2}\int {u}^{-1}du\hfill & =\frac{1}{2}\text{ln}|u|+C\hfill \\ \\ & =\frac{1}{2}\text{ln}|{x}^{4}+3{x}^{2}|+C.\hfill \end{array}[/latex]

Finding an Antiderivative of a Logarithmic Function

Find the antiderivative of the log function [latex]{\text{log}}_{2}x.[/latex]

Answer:

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

[latex]\int {\text{log}}_{2}xdx=\frac{x}{\text{ln}2}(\text{ln}x-1)+C.[/latex]

Find the antiderivative of [latex]{\text{log}}_{3}x.[/latex]

Answer:

[latex]\frac{x}{\text{ln}3}(\text{ln}x-1)+C[/latex]

Hint

Follow (Figure) and refer to the rule on integration formulas involving logarithmic functions.

Next, we will focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also in Derivatives, we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.

Integrals that Result in Inverse Sine Functions

Let us begin with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Rule: Integration Formulas Resulting in Inverse Trigonometric Functions

The following integration formulas yield inverse trigonometric functions:

  1. [latex]\int \frac{du}{\sqrt{{a}^{2}-{u}^{2}}}={ \sin }^{-1}\frac{u}{a}+C[/latex]
  2. [latex]\int \frac{du}{{a}^{2}+{u}^{2}}=\frac{1}{a}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}\frac{u}{a}+C[/latex]
  3. [latex]\int \frac{du}{u\sqrt{{u}^{2}-{a}^{2}}}=\frac{1}{a}\phantom{\rule{0.05em}{0ex}}{ \sec }^{-1}\frac{u}{a}+C[/latex]

Proof

Let [latex]y={ \sin }^{-1}\frac{x}{a}.[/latex] Then [latex]a \sin y=x.[/latex] Now let’s use implicit differentiation. We obtain

[latex]\begin{array}{ccc}\hfill \frac{d}{dx}(a \sin y)& =\hfill & \frac{d}{dx}(x)\hfill \\ \\ \hfill a \cos y\frac{dy}{dx}& =\hfill & 1\hfill \\ \hfill \frac{dy}{dx}& =\hfill & \frac{1}{a \cos y}.\hfill \end{array}[/latex]

For [latex]-\frac{\pi }{2}\le y\le \frac{\pi }{2}, \cos y\ge 0.[/latex] Thus, applying the Pythagorean identity [latex]{ \sin }^{2}y+{ \cos }^{2}y=1,[/latex] we have [latex] \cos y=\sqrt{1={ \sin }^{2}y}.[/latex] This gives

[latex]\begin{array}{cc}\frac{1}{a \cos y}\hfill & =\frac{1}{a\sqrt{1-{ \sin }^{2}y}}\hfill \\ \\ & =\frac{1}{\sqrt{{a}^{2}-{a}^{2}{ \sin }^{2}y}}\hfill \\ & =\frac{1}{\sqrt{{a}^{2}-{x}^{2}}}.\hfill \end{array}[/latex]

Then for [latex]\text{−}a\le x\le a,[/latex] we have

[latex]\int \frac{1}{\sqrt{{a}^{2}-{u}^{2}}}du={ \sin }^{-1}(\frac{u}{a})+C.[/latex]

□

Find the antiderivative of [latex]\int \frac{dx}{\sqrt{1-16{x}^{2}}}.[/latex]

Answer: [latex]\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{ \sin }^{-1}(4x)+C[/latex]

Hint

Substitute [latex]u=4x[/latex]

Finding an Antiderivative Involving an Inverse Trigonometric Function

Evaluate the integral [latex]\int \frac{dx}{\sqrt{4-9{x}^{2}}}.[/latex]

Answer: Substitute [latex]u=3x.[/latex] Then [latex]du=3dx[/latex] and we have

[latex]\int \frac{dx}{\sqrt{4-9{x}^{2}}}=\frac{1}{3}\int \frac{du}{\sqrt{4-{u}^{2}}}.[/latex]

Applying the formula with [latex]a=2,[/latex] we obtain

[latex]\begin{array}{cc}\int \frac{dx}{\sqrt{4-9{x}^{2}}}\hfill & =\frac{1}{3}\int \frac{du}{\sqrt{4-{u}^{2}}}\hfill \\ \\ & =\frac{1}{3}{ \sin }^{-1}(\frac{u}{2})+C\hfill \\ & =\frac{1}{3}{ \sin }^{-1}(\frac{3x}{2})+C.\hfill \end{array}[/latex]

Find the indefinite integral using an inverse trigonometric function and substitution for [latex]\int \frac{dx}{\sqrt{9-{x}^{2}}}.[/latex]

Answer:

[latex]{ \sin }^{-1}(\frac{x}{3})+C[/latex]

Hint

Use the formula in the rule on integration formulas resulting in inverse trigonometric functions.

Integrals Resulting in Other Inverse Trigonometric Functions

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.

Finding an Antiderivative Involving the Inverse Tangent Function

Find an antiderivative of [latex]\int \frac{1}{1+4{x}^{2}}dx.[/latex]

Answer:

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for [latex]{ \tan }^{-1}u+C.[/latex] So we use substitution, letting [latex]u=2x,[/latex] then [latex]du=2dx[/latex] and [latex]1\text{/}2du=dx.[/latex] Then, we have

[latex]\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}u+C=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(2x)+C.[/latex]

Use substitution to find the antiderivative of [latex]\int \frac{dx}{25+4{x}^{2}}.[/latex]

Answer:

[latex]\frac{1}{10}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(\frac{2x}{5})+C[/latex]

Hint

Use the solving strategy from (Figure) and the rule on integration formulas resulting in inverse trigonometric functions.

Applying the Integration Formulas

Find the antiderivative of [latex]\int \frac{1}{9+{x}^{2}}dx.[/latex]

Apply the formula with [latex]a=3.[/latex] Then,

[latex]\int \frac{dx}{9+{x}^{2}}=\frac{1}{3}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(\frac{x}{3})+C.[/latex]

Find the antiderivative of [latex]\int \frac{dx}{16+{x}^{2}}.[/latex]

Answer:

[latex]\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(\frac{x}{4})+C[/latex]

Hint

Follow the steps in (Figure).

Key Concepts

  • Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable [latex]u[/latex] and du for appropriate expressions in the integrand.
  • When using substitution for a definite integral, we also have to change the limits of integration.

Key Equations

  • Substitution with Indefinite Integrals [latex]\int f\left[g(x)\right]{g}^{\prime }(x)dx=\int f(u)du=F(u)+C=F(g(x))+C[/latex]
  • Substitution with Definite Integrals [latex]{\int }_{a}^{b}f(g(x))g\text{‘}(x)dx={\int }_{g(a)}^{g(b)}f(u)du[/latex]

1. Why is [latex]u[/latex]-substitution referred to as change of variable?

2. If [latex]f=g\circ h,[/latex] when reversing the chain rule, [latex]\frac{d}{dx}(g\circ h)(x)={g}^{\prime }(h(x)){h}^{\prime }(x),[/latex] should you take [latex]u=g(x)[/latex] or [latex]u=h(x)?[/latex]

Answer:

[latex]u=h(x)[/latex]

In the following exercises, verify each identity using differentiation. Then, using the indicated [latex]u[/latex]-substitution, identify [latex]f[/latex] such that the integral takes the form [latex]\int f(u)du.[/latex]

3. [latex]\int x\sqrt{x+1}dx=\frac{2}{15}{(x+1)}^{3\text{/}2}(3x-2)+C;u=x+1[/latex]

4. [latex]\int \frac{{x}^{2}}{\sqrt{x-1}}dx(x>1)=\frac{2}{15}\sqrt{x-1}(3{x}^{2}+4x+8)+C;u=x-1[/latex]

Answer:

[latex]f(u)=\frac{{(u+1)}^{2}}{\sqrt{u}}[/latex]

5. [latex]\int x\sqrt{4{x}^{2}+9}dx=\frac{1}{12}{(4{x}^{2}+9)}^{3\text{/}2}+C;u=4{x}^{2}+9[/latex]

6. [latex]\int \frac{x}{\sqrt{4{x}^{2}+9}}dx=\frac{1}{4}\sqrt{4{x}^{2}+9}+C;u=4{x}^{2}+9[/latex]

Answer:

[latex]du=8xdx;f(u)=\frac{1}{8\sqrt{u}}[/latex]

7. [latex]\int \frac{x}{{(4{x}^{2}+9)}^{2}}dx=-\frac{1}{8(4{x}^{2}+9)};u=4{x}^{2}+9[/latex]

In the following exercises, find the antiderivative using the indicated substitution.

8. [latex]\int {(x+1)}^{4}dx;u=x+1[/latex]

Answer:

[latex]\frac{1}{5}{(x+1)}^{5}+C[/latex]

9. [latex]\int {(x-1)}^{5}dx;u=x-1[/latex]

10. [latex]\int {(2x-3)}^{-7}dx;u=2x-3[/latex]

Answer:

[latex]-\frac{1}{12{(3-2x)}^{6}}+C[/latex]

11. [latex]\int {(3x-2)}^{-11}dx;u=3x-2[/latex]

12. [latex]\int \frac{x}{\sqrt{{x}^{2}+1}}dx;u={x}^{2}+1[/latex]

Answer:

[latex]\sqrt{{x}^{2}+1}+C[/latex]

13. [latex]\int \frac{x}{\sqrt{1-{x}^{2}}}dx;u=1-{x}^{2}[/latex]

14. [latex]\int (x-1){({x}^{2}-2x)}^{3}dx;u={x}^{2}-2x[/latex]

Answer:

[latex]\frac{1}{8}{({x}^{2}-2x)}^{4}+C[/latex]

15. [latex]\int ({x}^{2}-2x){({x}^{3}-3{x}^{2})}^{2}dx;u={x}^{3}=3{x}^{2}[/latex]

16. [latex]\int { \cos }^{3}\theta d\theta ;u= \sin \theta [/latex]([latex]Hint\text{:}{ \cos }^{2}\theta =1-{ \sin }^{2}\theta [/latex])

Answer:

[latex] \sin \theta -\frac{{ \sin }^{3}\theta }{3}+C[/latex]

17. [latex]\int { \sin }^{3}\theta d\theta ;u= \cos \theta [/latex][latex](Hint\text{:}{ \sin }^{2}\theta =1-{ \cos }^{2}\theta)[/latex]

In the following exercises, use a suitable change of variables to determine the indefinite integral.

18. [latex]\int x{(1-x)}^{99}dx[/latex]

Answer:

[latex]\frac{{(1-x)}^{101}}{101}-\frac{{(1-x)}^{100}}{100}+C[/latex]

19. [latex]\int t{(1-{t}^{2})}^{10}dt[/latex]

20. [latex]\int {(11x-7)}^{-3}dx[/latex]

Answer:

[latex]-\frac{1}{22(7-11{x}^{2})}+C[/latex]

21. [latex]\int {(7x-11)}^{4}dx[/latex]

22. [latex]\int { \cos }^{3}\theta \sin \theta d\theta [/latex]

Answer:

[latex]-\frac{{ \cos }^{4}\theta }{4}+C[/latex]

23. [latex]\int { \sin }^{7}\theta \cos \theta d\theta [/latex]

24. [latex]\int { \cos }^{2}(\pi t) \sin (\pi t)dt[/latex]

Answer:

[latex]-\frac{{ \cos }^{3}(\pi t)}{3\pi }+C[/latex]

25. [latex]\int { \sin }^{2}x{ \cos }^{3}xdx[/latex][latex](Hint\text{:}{ \sin }^{2}x+{ \cos }^{2}x=1)[/latex]

26. [latex]\int t \sin ({t}^{2}) \cos ({t}^{2})dt[/latex]

Answer:

[latex]-\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{ \cos }^{2}({t}^{2})+C[/latex]

27. [latex]\int {t}^{2}{ \cos }^{2}({t}^{3}) \sin ({t}^{3})dt[/latex]

28. [latex]\int \frac{{x}^{2}}{{({x}^{3}-3)}^{2}}dx[/latex]

Answer:

[latex]-\frac{1}{3({x}^{3}-3)}+C[/latex]

29. [latex]\int \frac{{x}^{3}}{\sqrt{1-{x}^{2}}}dx[/latex]

30. [latex]\int \frac{{y}^{5}}{{(1-{y}^{3})}^{3\text{/}2}}dy[/latex]

Answer:

[latex]-\frac{2({y}^{3}-2)}{3\sqrt{1-{y}^{3}}}[/latex]

31. [latex]{\int \cos \theta (1- \cos \theta )}^{99} \sin \theta d\theta [/latex]

32. [latex]{\int (1-{ \cos }^{3}\theta )}^{10}{ \cos }^{2}\theta \sin \theta d\theta [/latex]

Answer:

[latex]\frac{1}{33}{(1-{ \cos }^{3}\theta )}^{11}+C[/latex]

33. [latex]\int ( \cos \theta -1){({ \cos }^{2}\theta -2 \cos \theta )}^{3} \sin \theta d\theta [/latex]

34. [latex]\int ({ \sin }^{2}\theta -2 \sin \theta ){({ \sin }^{3}\theta -3{ \sin }^{2}\theta )}^{3} \cos \theta d\theta [/latex]

Answer:

[latex]\frac{1}{12}{({ \sin }^{3}\theta -3{ \sin }^{2}\theta )}^{4}+C[/latex]

35. [latex]\int {e}^{2x}dx[/latex]

36. [latex]\int {e}^{-3x}dx[/latex]

Answer:

[latex]\frac{-1}{3}{e}^{-3x}+C[/latex]

37. [latex]\int {2}^{x}dx[/latex]

38. [latex]\int {3}^{\text{−}x}dx[/latex]

Answer:

[latex]-\frac{{3}^{\text{−}x}}{\text{ln}3}+C[/latex]

39. [latex]\int \frac{1}{2x}dx[/latex]

40. [latex]\int \frac{2}{x}dx[/latex]

Answer:

[latex]\text{ln}({x}^{2})+C[/latex]

41. [latex]\int \frac{1}{{x}^{2}}dx[/latex]

42. [latex]\int \frac{1}{\sqrt{x}}dx[/latex]

Answer:

[latex]2\sqrt{x}+C[/latex]

In the following exercises, find each indefinite integral by using appropriate substitutions.

43. [latex]\int \frac{\text{ln}x}{x}dx[/latex]

44. [latex]\int \frac{dx}{x{(\text{ln}x)}^{2}}[/latex]

Answer:

[latex]-\frac{1}{\text{ln}x}+C[/latex]

45. [latex]\int \frac{dx}{x\text{ln}x}(x>1)[/latex]

46. [latex]\int \frac{dx}{x\text{ln}x\text{ln}(\text{ln}x)}[/latex]

Answer:

[latex]\text{ln}(\text{ln}(\text{ln}x))+C[/latex]

47. [latex]\int \tan \theta d\theta [/latex]

48. [latex]\int \frac{ \cos x-x \sin x}{x \cos x}dx[/latex]

Answer: [latex]\text{ln}(x \cos x)+C[/latex]

49. [latex]\int \frac{\text{ln}( \sin x)}{ \tan x}dx[/latex]

50. [latex]\int \text{ln}( \cos x) \tan xdx[/latex]

Answer: [latex]-\frac{1}{2}{(\text{ln}( \cos (x)))}^{2}+C[/latex]

51. [latex]\int x{e}^{\text{−}{x}^{2}}dx[/latex]

52. [latex]\int {x}^{2}{e}^{\text{−}{x}^{3}}dx[/latex]

Answer:

[latex]\frac{\text{−}{e}^{\text{−}{x}^{3}}}{3}+C[/latex]

53. [latex]\int {e}^{ \sin x} \cos xdx[/latex]

54. [latex]\int {e}^{ \tan x}{ \sec }^{2}xdx[/latex]

Answer:

[latex]{e}^{ \tan x}+C[/latex]

55. [latex]\int {e}^{\text{ln}x}\frac{dx}{x}[/latex]

56. [latex]\int \frac{{e}^{\text{ln}(1-t)}}{1-t}dt[/latex]

Answer:

[latex]t+C[/latex]

In the following exercises, verify by differentiation that [latex]\int \text{ln}xdx=x(\text{ln}x-1)+C,[/latex] then use appropriate changes of variables to compute the integral.
57. [latex]\int \text{ln}xdx[/latex][latex](Hint\text{:}\int \text{ln}xdx=\frac{1}{2}\int x\text{ln}({x}^{2})dx)[/latex]

58. [latex]\int {x}^{2}{\text{ln}}^{2}xdx[/latex]

Answer:

[latex]\frac{1}{9}{x}^{3}(\text{ln}({x}^{3})-1)+C[/latex]

59. [latex]\int \frac{\text{ln}x}{{x}^{2}}dx[/latex][latex](Hint\text{:}\text{Set}u=\frac{1}{x}\text{.})[/latex]

60. [latex]\int \frac{\text{ln}x}{\sqrt{x}}dx[/latex][latex](Hint\text{:}\text{Set}u=\sqrt{x}\text{.})[/latex]

Answer:

[latex]2\sqrt{x}(\text{ln}x-2)+C[/latex]

61. Write an integral to express the area under the graph of [latex]y=\frac{1}{t}[/latex] from [latex]t=1[/latex] to ex and evaluate the integral.

62. Write an integral to express the area under the graph of [latex]y={e}^{t}[/latex] between [latex]t=0[/latex] and [latex]t=\text{ln}x,[/latex] and evaluate the integral.

Answer:

[latex]{\int }_{0}^{\text{ln}x}{e}^{t}dt={e}^{t}{|}_{0}^{\text{ln}x}={e}^{\text{ln}x}-{e}^{0}=x-1[/latex]

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

63. [latex]\int \tan (2x)dx[/latex]

64. [latex]\int \frac{ \sin (3x)- \cos (3x)}{ \sin (3x)+ \cos (3x)}dx[/latex]

Answer:

[latex]-\frac{1}{3}\text{ln}( \sin (3x)+ \cos (3x))[/latex]

65. [latex]\int \frac{x \sin ({x}^{2})}{ \cos ({x}^{2})}dx[/latex]

66. [latex]\int x \csc ({x}^{2})dx[/latex]

Answer:

[latex]-\frac{1}{2}\text{ln}| \csc ({x}^{2})+ \cot ({x}^{2})|+C[/latex]

67. [latex]\int \text{ln}( \cos x) \tan xdx[/latex]

68. [latex]\int \text{ln}( \csc x) \cot xdx[/latex]

Answer:

[latex]-\frac{1}{2}{(\text{ln}( \csc x))}^{2}+C[/latex]

69. [latex]\int \frac{{e}^{x}-{e}^{\text{−}x}}{{e}^{x}+{e}^{\text{−}x}}dx[/latex]

In the following exercises, evaluate the definite integral.

70. [latex]{\int }_{1}^{2}\frac{1+2x+{x}^{2}}{3x+3{x}^{2}+{x}^{3}}dx[/latex]

Answer: [latex]\frac{1}{3}\text{ln}(\frac{26}{7})[/latex]

In the following exercises, integrate using the indicated substitution.

71. [latex]\int \frac{x}{x-100}dx;u=x-100[/latex]

72. [latex]\int \frac{y-1}{y+1}dy;u=y+1[/latex]

Answer:

[latex]y-2\text{ln}|y+1|+C[/latex]

73. [latex]\int \frac{1-{x}^{2}}{3x-{x}^{3}}dx;u=3x-{x}^{3}[/latex]

74. [latex]\int \frac{ \sin x+ \cos x}{ \sin x- \cos x}dx;u= \sin x- \cos x[/latex]

Answer:

[latex]\text{ln}| \sin x- \cos x|+C[/latex]

75. [latex]\int {e}^{2x}\sqrt{1-{e}^{2x}}dx;u={e}^{2x}[/latex]

76. [latex]\int \text{ln}(x)\frac{\sqrt{1-{(\text{ln}x)}^{2}}}{x}dx;u=\text{ln}x[/latex]

Answer:

[latex]-\frac{1}{3}{(1-(\text{ln}{x}^{2}))}^{3\text{/}2}+C[/latex]

In the following exercises, find each indefinite integral, using appropriate substitutions.

77. [latex]\int \frac{dx}{\sqrt{9-{x}^{2}}}[/latex]

Answer: [latex]{ \sin }^{-1}(\frac{x}{3})+C[/latex]

78. [latex]\int \frac{dx}{\sqrt{1-16{x}^{2}}}[/latex]

79. [latex]\int \frac{dx}{9+{x}^{2}}[/latex]

Answer:

[latex]\frac{1}{3}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(\frac{x}{3})+C[/latex]

80. [latex]\int \frac{dx}{25+16{x}^{2}}[/latex]

81. [latex]\int \frac{dx}{|x|\sqrt{{x}^{2}-9}}[/latex]

Answer:

[latex]\frac{1}{3}\phantom{\rule{0.05em}{0ex}}{ \sec }^{-1}(\frac{x}{3})+C[/latex]

82. [latex]\int \frac{dx}{|x|\sqrt{4{x}^{2}-16}}[/latex]

83. Explain the relationship [latex]\text{−}{ \cos }^{-1}t+C=\int \frac{dt}{\sqrt{1-{t}^{2}}}={ \sin }^{-1}t+C.[/latex] Is it true, in general, that [latex]{ \cos }^{-1}t=\text{−}{ \sin }^{-1}t?[/latex]

Answer:

[latex] \cos (\frac{\pi }{2}-\theta )= \sin \theta .[/latex] So, [latex]{ \sin }^{-1}t=\frac{\pi }{2}-{ \cos }^{-1}t.[/latex] They differ by a constant.

84. Explain the relationship [latex]{ \sec }^{-1}t+C=\int \frac{dt}{|t|\sqrt{{t}^{2}-1}}=\text{−}{ \csc }^{-1}t+C.[/latex] Is it true, in general, that [latex]{ \sec }^{-1}t=\text{−}{ \csc }^{-1}t?[/latex]

In the following exercises, compute the antiderivative using appropriate substitutions.

85. [latex]\int \frac{{ \sin }^{-1}tdt}{\sqrt{1-{t}^{2}}}[/latex]

Answer:

[latex]\frac{1}{2}{({ \sin }^{-1}t)}^{2}+C[/latex]

86. [latex]\int \frac{dt}{{ \sin }^{-1}t\sqrt{1-{t}^{2}}}[/latex]

87. [latex]\int \frac{{ \tan }^{-1}(2t)}{1+4{t}^{2}}dt[/latex]

Answer: [latex]\frac{1}{4}{({ \tan }^{-1}(2t))}^{2}[/latex]

88. [latex]\int \frac{t{ \tan }^{-1}({t}^{2})}{1+{t}^{4}}dt[/latex]

89. [latex]\int \frac{{ \sec }^{-1}(\frac{t}{2})}{|t|\sqrt{{t}^{2}-4}}dt[/latex]

Answer:

[latex]\frac{1}{4}({ \sec }^{-1}{(\frac{t}{2})}^{2})+C[/latex]

90. [latex]\int \frac{t{ \sec }^{-1}({t}^{2})}{{t}^{2}\sqrt{{t}^{4}-1}}dt[/latex]

In the following exercises, compute each integral using appropriate substitutions.

91. [latex]\int \frac{{e}^{x}}{\sqrt{1-{e}^{2t}}}dt[/latex]

Answer:

[latex]{ \sin }^{-1}({e}^{t})+C[/latex]

92. [latex]\int \frac{{e}^{t}}{1+{e}^{2t}}dt[/latex]

93. [latex]\int \frac{dt}{t\sqrt{1-{\text{ln}}^{2}t}}[/latex]

Answer:

[latex]{ \sin }^{-1}(\text{ln}t)+C[/latex]

94. [latex]\int \frac{dt}{t(1+{\text{ln}}^{2}t)}[/latex]

95. [latex]\int \frac{{ \cos }^{-1}(2t)}{\sqrt{1-4{t}^{2}}}dt[/latex]

Answer:

[latex]-\frac{1}{2}{({ \cos }^{-1}(2t))}^{2}+C[/latex]

96. [latex]\int \frac{{e}^{t}{ \cos }^{-1}({e}^{t})}{\sqrt{1-{e}^{2t}}}dt[/latex]

97. For [latex]1<B<\infty ,[/latex] compute [latex]I(B)={\int }_{1}^{B}\frac{dt}{t\sqrt{{t}^{2}-1}}[/latex] and evaluate [latex]\underset{B\to \infty }{\text{lim}}I(B),[/latex] the area under the graph of [latex]\frac{1}{t\sqrt{{t}^{2}-1}}[/latex] over [latex][1,\infty ).[/latex]

98. Use the substitution [latex]u=\sqrt{2} \cot x[/latex] and the identity [latex]1+{ \cot }^{2}x={ \csc }^{2}x[/latex] to evaluate [latex]\int \frac{dx}{1+{ \cos }^{2}x}.[/latex] (Hint: Multiply the top and bottom of the integrand by [latex]{ \csc }^{2}x.[/latex])

Answer:

Using the hint, one has [latex]\int \frac{{ \csc }^{2}x}{{ \csc }^{2}x+{ \cot }^{2}x}dx=\int \frac{{ \csc }^{2}x}{1+2{ \cot }^{2}x}dx.[/latex] Set [latex]u=\sqrt{2} \cot x.[/latex] Then, [latex]du=\text{−}\sqrt{2}{ \csc }^{2}x[/latex] and the integral is [latex]-\frac{1}{\sqrt{2}}\int \frac{du}{1+{u}^{2}}=-\frac{1}{\sqrt{2}}{ \tan }^{-1}u+C=\frac{1}{\sqrt{2}}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(\sqrt{2} \cot x)+C.[/latex] If one uses the identity [latex]{ \tan }^{-1}s+{ \tan }^{-1}(\frac{1}{s})=\frac{\pi }{2},[/latex] then this can also be written [latex]\frac{1}{\sqrt{2}}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(\frac{ \tan x}{\sqrt{2}})+C.[/latex]

99. [T] Approximate the points at which the graphs of [latex]f(x)=2{x}^{2}-1[/latex] and [latex]g(x)={(1+4{x}^{2})}^{-3\text{/}2}[/latex] intersect, and approximate the area between their graphs accurate to three decimal places.

100. [T] Approximate the points at which the graphs of [latex]f(x)={x}^{2}-1[/latex] and [latex]f(x)={x}^{2}-1[/latex] intersect, and approximate the area between their graphs accurate to three decimal places.

Answer:

[latex]x\approx ±1.13525.[/latex] The left endpoint estimate with [latex]N=100[/latex] is 2.796 and these decimals persist for [latex]N=500.[/latex]

Glossary

change of variables
the substitution of a variable, such as [latex]u[/latex], for an expression in the integrand
integration by substitution
a technique for integration that allows integration of functions that are the result of a chain-rule derivative