# 4.1: Integration by Substitution

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We motivate this section with an example. Let \(f(x) = (x^2+3x-5)^{10}\). We can compute \(f'(x)\) using the Chain Rule. It is:

$$f'(x) = 10(x^2+3x-5)^9\cdot(2x+3) = (20x+30)(x^2+3x-5)^9.$$

Now consider this: What is \(\int (20x+30)(x^2+3x-5)^9\ dx\)? We have the answer in front of us;

$$\int (20x+30)(x^2+3x-5)^9\ dx = (x^2+3x-5)^{10}+C.$$

How would we have evaluated this indefinite integral without starting with \(f(x)\) as we did?

This section explores *integration by substitution.* It allows us to "undo the Chain Rule." Substitution allows us to evaluate the above integral without knowing the original function first.

The underlying principle is to rewrite a "complicated" integral of the form \(\int f(x)\ dx\) as a not--so--complicated integral \(\int h(u)\ du\). We'll formally establish later how this is done. First, consider again our introductory indefinite integral, \(\int (20x+30)(x^2+3x-5)^9\ dx\). Arguably the most "complicated" part of the integrand is \((x^2+3x-5)^9\). We wish to make this simpler; we do so through a substitution. Let \(u=x^2+3x-5\). Thus

$$(x^2+3x-5)^9 = u^9.$$

We have established \(u\) as a function of \(x\), so now consider the differential of \(u\):

$$du = (2x+3)dx.$$

Keep in mind that \((2x+3)\) and \(dx\) are multiplied; the \(dx\) is not "just sitting there."

Return to the original integral and do some substitutions through algebra:

\[\begin{align} \int (20x+30)(x^2+3x-5)^9\ dx &= \int 10(2x+3)(x^2+3x-5)^9\ dx \\ &=\int 10(\underbrace{x^2+3x-5}_u)^9\underbrace{(2x+3)\ dx}_{du} \\ &=\int 10u^9\ du \\ &= u^{10} + C \quad \text{(replace \(u\) with \(x^2+3x-5\))}\\ &= (x^2+3x-5)^{10} +C\end{align}\]

One might well look at this and think "I (sort of) followed how that worked, but I could never come up with that on my own," but the process is learnable. This section contains numerous examples through which the reader will gain understanding and mathematical maturity enabling them to regard substitution as a natural tool when evaluating integrals.

We stated before that integration by substitution "undoes" the Chain Rule. Specifically, let \(F(x)\) and \(g(x)\) be differentiable functions and consider the derivative of their composition:

$$\frac{d}{dx}\Big(F\big(g(x)\big)\Big) = F'(g(x))g'(x).$$

Thus

$$\int F'(g(x))g'(x)\ dx = F(g(x)) + C.$$

Integration by substitution works by recognizing the "inside" function \(g(x)\) and replacing it with a variable. By setting \(u=g(x)\), we can rewrite the derivative as

$$\frac{d}{dx}\Big(F\big(u\big)\Big) = F'(u)u'.$$

Since \(du = g'(x)dx\), we can rewrite the above integral as

$$\int F'(g(x))g'(x)\ dx = \int F'(u) du = F(u)+C = F(g(x))+ C.$$

This concept is important so we restate it in the context of a theorem.

Theorem \(\PageIndex{1}\): Integration by Substitution

Let \(F\) and \(g\) be differentiable functions, where the range of \(g\) is an interval \(I\) contained in the domain of \(F\). Then

$$\int F'(g(x))g'(x)\ dx = F(g(x)) + C.$$

If \(u = g(x)\), then \(du = g'(x)dx\) and

$$\int F'(g(x))g'(x)\ dx = \int F'(u)\ du = F(u)+C = F(g(x))+C.$$

The point of substitution is to make the integration step easy. Indeed, the step \(\int F'(u)\ du = F(u) + C\) looks easy, as the antiderivative of the derivative of \(F\) is just \(F\), plus a constant. The "work" involved is making the proper substitution. There is not a step-by-step process that one can memorize; rather, experience will be one's guide. To gain experience, we now embark on many examples.

Example \(\PageIndex{1}\): Integrating by substitution

Evaluate \(\int x\sin(x^2+5)\ dx\).

**Solution**

Knowing that substitution is related to the Chain Rule, we choose to let \(u\) be the "inside" function of \(\sin(x^2+5)\). (This is not *always* a good choice, but it is often the best place to start.)

Let \(u = x^2+5\), hence \(du = 2x\,dx\). The integrand has an \(x\,dx\) term, but not a \(2x\,dx\) term. (Recall that multiplication is commutative, so the \(x\) does not physically have to be next to \(dx\) for there to be an \(x\,dx\) term.) We can divide both sides of the \(du\) expression by 2:

$$du = 2x\,dx \quad \Rightarrow \quad \frac12du = x\,dx.$$

We can now substitute.

\[\begin{align}\int x\sin(x^2+5)\ dx &= \int \sin(\underbrace{x^2+5}_u) \underbrace{x\ dx}_{\frac12du}\\ &= \int \frac12\sin u\ du\end{align}\]

\[\begin{align} \phantom{\int x\sin(x^2+5)\ dx} &= -\frac12\cos u + C \quad \text{ (now replace \(u\) with \(x^2+5\))}\\ &=-\frac12\cos(x^2+5) + C. \end{align}\]

Thus \(\int x\sin(x^2+5)\ dx = -\frac12\cos(x^2+5)+C\). We can check our work by evaluating the derivative of the right hand side.

Exercise \(\PageIndex{1}\)

Evaluate \(\int x\cos(x^2+5)\ dx\).

**Answer**-
\frac12\sin(x^2+5)+C.

Example \(\PageIndex{2}\): Integrating by substitution

Evaluate \(\int \cos(5x)\ dx\).

**Solution**

Again let \(u\) replace the "inside" function. Letting \(u = 5x\), we have \(du = 5dx\). Since our integrand does not have a \(5dx\) term, we can divide the previous equation by \(5\) to obtain \(\frac15du = dx\). We can now substitute.

\[\begin{align} \int \cos(5x)\ dx &= \int \cos(\underbrace{5x}_u) \underbrace{dx}_{\frac{1}{5}du} \\&= \int \frac{1}{5}\cos u \ du \\ &= \frac{1}{5} \sin u + C \\&= \frac{1}{5}\sin (5x)+C. \end{align}\]

We can again check our work through differentiation.

Exercise \(\PageIndex{2}\)

Evaluate \(\int \sin(5x)\ dx\).

**Answer**-
-\frac{1}{5}\cos(5x)+C. .

The previous example exhibited a common, and simple, type of substitution. The "inside" function was a linear function (in this case, \(y = 5x\)). When the inside function is linear, the resulting integration is very predictable, outlined here.

Key Idea 10: Substitution With A Linear Function

Consider \(\int F'(ax+b)\ dx\), where \(a\neq 0\) and \(b\) are constants. Letting \(u = ax+b\) gives \(du = a\cdot dx\), leading to the result

$$\int F'(ax+b)\ dx = \frac{1}{a}F(ax+b) + C.$$

Thus \(\int \sin (7x-4)\ dx = -\frac17\cos(7x-4)+C\). Our next example can use Key Idea 10, but we will only employ it after going through all of the steps.

Example \(\PageIndex{3}\): Integrating by substituting a linear function

Evaluate \(\int \frac{7}{-3x+1}\ dx\).

**Solution**

View this a composition of functions \(f(g(x))\), where \(f(x) = 7/x\) and \(g(x) = -3x+1\). Employing our understanding of substitution, we let \(u = -3x+1\), the inside function. Thus \(du = -3dx\). The integrand lacks a \(-3\); hence divide the previous equation by \(-3\) to obtain \(-du/3 = dx\). We can now evaluate the integral through substitution.

\[\begin{align} \int \frac{7}{-3x+1}\ dx &= \int \frac{7}{u}\frac{du}{-3} \\ &= \frac{-7}{3}\int \frac{du}{u} \\ &= \frac{-7}{3}\ln |u| + C\\ &=-\frac{7}{3}\ln|-3x+1| + C. \end{align}\]

Using Key Idea 10 is faster, recognizing that \(u\) is linear and \(a = -3\). One may want to continue writing out all the steps until they are comfortable with this particular shortcut.

Exercise \(\PageIndex{3}\)

Evaluate \(\int \frac{7}{3x+1}\ dx\).

**Answer**-
\frac{7}{3|\ln|3x+1| + C.

Not all integrals that benefit from substitution have a clear "inside" function. Several of the following examples will demonstrate ways in which this occurs.

Example \(\PageIndex{4}\): Integrating by substitution

Evaluate \(\int \sin x\cos x\ dx\).

**Solution**

There is not a composition of function here to exploit; rather, just a product of functions. Do not be afraid to experiment; when given an integral to evaluate, it is often beneficial to think "If I let \(u\) be *this*, then \(du\) must be *that* ..." and see if this helps simplify the integral at all.

In this example, let's set \(u = \sin x\). Then \(du = \cos x\ dx\), which we have as part of the integrand! The substitution becomes very straightforward:

\[ \begin{align} \int \sin x\cos x\ dx &= \int u\ du \\ &= \frac{1}{2}u^2+ C \\ &= \frac{1}{2}\sin^2 x + C. \end{align}\]

One would do well to ask "What would happen if we let \(u = \cos x\)?" The result is just as easy to find, yet looks very different. The challenge to the reader is to evaluate the integral letting \(u = \cos x\) and discover why the answer is the same, yet looks different.

Exercise \(\PageIndex{4}\)

Evaluate \(\int \sin^3 x\cos x\ dx\)..

**Answer**-
\frac{1}{3}\sin^4 x + C.

Our examples so far have required "basic substitution." The next example demonstrates how substitutions can be made that often strike the new learner as being "nonstandard."

Example \(\PageIndex{5}\): Integrating by substitution

Evaluate \(\int x\sqrt{x+3}\ dx\).

**Solution**

Recognizing the composition of functions, set \(u = x+3\). Then \(du = dx\), giving what seems initially to be a simple substitution. But at this stage, we have:

$$\int x\sqrt{x+3}\ dx = \int x\sqrt{u}\ du.$$

We cannot evaluate an integral that has both an \(x\) and an \(u\) in it. We need to convert the \(x\) to an expression involving just \(u\).

Since we set \(u = x+3\), we can also state that \(u-3 = x\). Thus we can replace \(x\) in the integrand with \(u-3\). It will also be helpful to rewrite \(\sqrt{u}\) as \(u^\frac12\).

\[\begin{align} \int x\sqrt{x+3} \ dx &= \int (u-3)u^\frac12\ du \\ &= \int \big(u^\frac32 - 3u^\frac12\big) \ du \\ &= \frac25u^\frac52 - 2u^\frac32 + C \\ &= \frac25(x+3)^\frac52 - 2(x+3)^\frac32 + C.\end{align}\]

Checking your work is always a good idea. In this particular case, some algebra will be needed to make one's answer match the integrand in the original problem.

Example \(\PageIndex{6}\): Integrating by substitution

Evaluate \(\int \frac{1}{x\ln x}\ dx\).

**Solution**

This is another example where there does not seem to be an obvious composition of functions. The line of thinking used in Example \(\PageIndex{5}\) is useful here: choose something for \(u\) and consider what this implies \(du\) must be. If \(u\) can be chosen such that \(du\) also appears in the integrand, then we have chosen well.

Choosing \(u = 1/x\) makes \(du = -1/x^2\ dx\); that does not seem helpful. However, setting \(u = \ln x\) makes \(du = 1/x\ dx\), which is part of the integrand. Thus:

\[\begin{align} \int \frac1{x\ln x}\ dx &= \int \frac{1}{\underbrace{\ln x}_{1/u}}\underbrace{\frac1x\ dx}_{du} \\ &= \int \frac1u\ du \\ &= \ln |u| + C \\ &= \ln | \ln x| + C.\end{align}\]

The final answer is interesting; the natural log of the natural log. Take the derivative to confirm this answer is indeed correct.

### Integrals Involving Trigonometric Functions

Section 6.3 delves deeper into integrals of a variety of trigonometric functions; here we use substitution to establish a foundation that we will build upon.

The next three examples will help fill in some missing pieces of our antiderivative knowledge. We know the antiderivatives of the sine and cosine functions; what about the other standard functions tangent, cotangent, secant and cosecant? We discover these next.

Example \(\PageIndex{7}\): Integration by substitution: antiderivatives of \(\tan x\)

Evaluate \(\int \tan x\ dx.\)

**Solution**

The previous paragraph established that we did not know the antiderivatives of tangent, hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral.

Rewrite \(\tan x\) as \(\sin x/\cos x\). While the presence of a composition of functions may not be immediately obvious, recognize that \(\cos x\) is "inside" the \(1/x\) function. Therefore, we see if setting \(u = \cos x\) returns usable results. We have that \(du = -\sin x\ dx\), hence \(-du = \sin x\ dx\). We can integrate:

\[\begin{align}\int \tan x \ dx &= \int \frac{\sin x}{\cos x}\ dx \\ &= \int \frac1{\underbrace{\cos x}_u}\underbrace{\sin x\ dx}_{-du} \\ &= \int \frac {-1}u \ du\\ &= -\ln |u| + C \\ &= -\ln |\cos x| + C.\end{align}\]

Some texts prefer to bring the \(-1\) inside the logarithm as a power of \(\cos x\), as in:

\[\begin{align} -\ln |\cos x| + C &= \ln |(\cos x)^{-1}| + C\\ &= \ln \left| \frac{1}{\cos x}\right| + C\\&= \ln |\sec x| + C.\end{align}\]

Thus the result they give is \(\int \tan x \ dx = \ln|\sec x| + C\). These two answers are equivalent.

Example \(\PageIndex{8}\): Integrating by substitution: antiderivatives of \(\sec x\)

Evaluate \(\int \sec x\ dx\).

**Solution**

This example employs a wonderful trick: multiply the integrand by "1" so that we see how to integrate more clearly. In this case, we write "1" as

$$1 = \frac{\sec x + \tan x}{\sec x + \tan x}.$$

This may seem like it came out of left field, but it works beautifully. Consider:

\[\begin{align} \int \sec x\ dx &= \int \sec x\cdot \frac{\sec x + \tan x}{\sec x + \tan x}\ dx \\ &= \int \frac{\sec^2 x + \sec x\tan x}{\sec x + \tan x}\ dx.\end{align}\]

Now let \(u = \sec x+\tan x\); this means \(du = (\sec x\tan x+ \sec^2 x)\ dx\), which is our numerator. Thus:

\[\begin{align} &= \int \frac{du}{u} \\ &= \ln |u| + C \\ &= \ln |\sec x+\tan x| + C.\end{align}\]

We can use similar techniques to those used in Examples \(\PageIndex{6}\) and \(\PageIndex{7}\) to find antiderivatives of \(\cot x\) and \(\csc x\) (which the reader can explore in the exercises.) We summarize our results here.

Theorem \(\PageIndex{1}\): Antiderivatives of Trigonometric Functions

- \( \int \sin x \ dx = -\cos x +C\)
- \(\int \cos x\ dx = \sin x + C\)
- \(\int \tan x\ dx = -\ln|\cos x|+C\)
- \( \int \csc x \ dx = -\ln|\csc x+\cot x| +C\)
- \(\int \sec x\ dx = \ln|\sec x+\tan x| + C\)
- \( \int \cot x\ dx = \ln|\sin x|+C\)

We explore one more common trigonometric integral.

Example \(\PageIndex{9}\): Integration by substitution: powers of \(\cos x\) and \(\sin x\)

Evaluate \(\int \cos^2x\ dx\).

**Solution**

We have a composition of functions as \(\cos^2x = \big(\cos x\big)^2\).

However, setting \(u = \cos x\) means \(du = -\sin x\ dx\), which we do not have in the integral. Another technique is needed.

The process we'll employ is to use a Power Reducing formula for \(\cos^2x\) (perhaps consult the back of this text for this formula), which states

$$\cos ^2x = \frac{1+\cos(2x)}{2}.$$

The right hand side of this equation is not difficult to integrate. We have:

\[\begin{align} \int \cos^2x\ dx &= \int \frac{1+\cos(2x)}2\ dx \\ &= \int \left( \frac12 + \frac12\cos(2x)\right)\ dx. \end{align} \]

Now use Key Idea 10:

\[\begin{align} &= \frac12x + \frac12\frac{\sin(2x)}{2} + C\\&= \frac12x + \frac{\sin(2x)}4 + C.\end{align}\]

We'll make significant use of this power--reducing technique in future sections.

### Simplifying the Integrand

It is common to be reluctant to manipulate the integrand of an integral; at first, our grasp of integration is tenuous and one may think that working with the integrand will improperly change the results. Integration by substitution works using a different logic: as long as *equality* is maintained, the integrand can be manipulated so that its *form* is easier to deal with. The next two examples demonstrate common ways in which using algebra first makes the integration easier to perform.

Example \(\PageIndex{10}\): Integration by substitution: simplifying first

Evaluate \(\int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\ dx\).

**Solution**

One may try to start by setting \(u\) equal to either the numerator or denominator; in each instance, the result is not workable.

When dealing with rational functions (i.e., quotients made up of polynomial functions), it is an almost universal rule that everything works better when the degree of the numerator is less than the degree of the denominator. Hence we use polynomial division.

We skip the specifics of the steps, but note that when \(x^2+2x+1\) is divided into \(x^3+4x^2+8x+5\), it goes in \(x+2\) times with a remainder of \(3x+3\). Thus

$$\frac{x^3+4x^2+8x+5}{x^2+2x+1} = x+2 + \frac{3x+3}{x^2+2x+1}.$$

Integrating \(x+2\) is simple. The fraction can be integrated by setting \(u = x^2+2x+1\), giving \(du = (2x+2)\ dx\). This is very similar to the numerator. Note that \(du/2 = (x+1)\ dx\) and then consider the following:

\[\begin{align}\int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\ dx & = \int \left(x+2 + \frac{3x+3}{x^2+2x+1}\right)\ dx \\ &= \int (x+2)\ dx + \int \frac{3(x+1)}{x^2+2x+1}\ dx \\ & = \frac12x^2+2x+C_1 + \int \frac{3}{u}\frac{du}{2} \\ &= \frac12x^2+2x+C_1 + \frac32\ln|u| + C_2 \\&= \frac12x^2+2x+\frac32\ln|x^2+2x+1| + C.\end{align}\]

In some ways, we "lucked out" in that after dividing, substitution was able to be done. In later sections we'll develop techniques for handling rational functions where substitution is not directly feasible.

Example \(\PageIndex{11}\): Integration by alternate methods

Evaluate \(\int \frac{x^2+2x+3}{\sqrt{x}}\ dx\) with, and without, substitution.

**Solution**

We already know how to integrate this particular example. Rewrite \(\sqrt{x}\) as \(x^\frac12\) and simplify the fraction:

$$ \frac{x^2+2x+3}{x^{1/2}} = x^\frac32 + 2x^\frac12 + 3x^{-\frac12}.$$

We can now integrate using the Power Rule:

\[\begin{align} \int \frac{x^2+2x+3}{x^{1/2}}\ dx &= \int\left(x^\frac32 + 2x^\frac12 + 3x^{-\frac12}\right)\ dx\\ &= \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12 + C\end{align}\]

This is a perfectly fine approach. We demonstrate how this can also be solved using substitution as its implementation is rather clever.

Let \(u = \sqrt{x} = x^\frac12\); therefore

$$du = \frac12x^{-\frac12}dx = \frac{1}{2\sqrt{x}}\ dx \quad \Rightarrow \quad 2du = \frac{1}{\sqrt{x}}\ dx.$$

This gives us \(\int \frac{x^2+2x+3}{\sqrt{x}}\ dx = \int (x^2+2x+3)\cdot2\ du\). What are we to do with the other \(x\) terms? Since \(u = x^\frac12\), \(u^2 = x\), etc. We can then replace \(x^2\) and \(x\) with appropriate powers of \(u\). We thus have

\[\begin{align*}\int \frac{x^2+2x+3}{\sqrt{x}}\ dx &= \int (x^2+2x+3)\cdot2\ du\\ &= \int 2(u^4 + 2u^2 + 3)\ du \\ &= \frac25u^5 + \frac43u^3 + 6u + C \\&= \frac25x^\frac52 + \frac43x^\frac32 + 6x^\frac12+C,\end{align*}\]

which is obviously the same answer we obtained before. In this situation, substitution is arguably more work than our other method. The fantastic thing is that it works. It demonstrates how flexible integration is.

### Substitution and Inverse Trigonometric Functions

When studying derivatives of inverse functions, we learned that

$$\frac{d}{dx}\big(\tan^{-1}x\big) = \frac{1}{1+x^2}.$$

Applying the Chain Rule to this is not difficult; for instance,

$$\frac{d}{dx}\big(\tan^{-1}5x\big) = \frac{5}{1+25x^2}.$$

We now explore how Substitution can be used to "undo" certain derivatives that are the result of the Chain Rule applied to Inverse Trigonometric functions. We begin with an example.

Example \(\PageIndex{12}\): Integrating by substitution: inverse trigonometric functions

Evaluate \( \int \frac{1}{25+x^2}\ dx\).

**Solution**

The integrand looks similar to the derivative of the arctangent function. Note:

\[\begin{align}\frac{1}{25+x^2} &= \frac{1}{25(1+\frac{x^2}{25})}\\ &= \frac{1}{25(1+\left(\frac{x}{5}\right)^2)} \\

&= \frac{1}{25}\frac{1}{1+\left(\frac{x}{5}\right)^2}\ .\end{align}\]

Thus

$$\int\frac{1}{25+x^2}\ dx = \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\ dx.$$

This can be integrated using Substitution. Set \(u = x/5\), hence \(du = dx/5\) or \(dx=5du\). Thus

\[\begin{align}\int\frac{1}{25+x^2}\ dx &= \frac{1}{25}\int \frac{1}{1+\left(\frac{x}{5}\right)^2}\ dx \\ &= \frac15\int \frac{1}{1+u^2}\ du \\ &= \frac15\tan^{-1}u + C \\ &= \frac15\tan^{-1}\left(\frac x5\right)+C\end{align}\]

Example \(\PageIndex{12}\) demonstrates a general technique that can be applied to other integrands that result in inverse trigonometric functions. The results are summarized here.

Theorem \(\PageIndex{2}\): Integrals Involving Inverse Trigonomentric Functions

Let \(a>0\).

- \( \int \frac{1}{a^2+x^2}\ dx = \frac1a\tan^{-1}\left(\frac{x}{a}\right) + C\)
- \( \int \frac{1}{\sqrt{a^2-x^2}}\ dx = \sin^{-1}\left(\frac{x}{a}\right)+C\)
- \( \int \frac{1}{x\sqrt{x^2-a^2}}\ dx = \frac1a\sec^{-1}\left(\frac{|x|}{a}\right)+C\)

Let's practice using Theorem \(\PageIndex{2}\).

Example \(\PageIndex{13}\): Integrating by substitution: inverse trigonometric functions

Evaluate the given indefinite integrals.

$$\int \frac{1}{9+x^2}\ dx,\quad \int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\ dx\quad \text{ and }\quad \int \frac{1}{\sqrt{5-x^2}}\ dx.$$

**Solution**

Each can be answered using a straightforward application of Theorem \(\PageIndex{2}\).

\(\int \frac{1}{9+x^2}\ dx = \frac13\tan^{-1} \frac x3 + C\), as \(a = 3\).

\(\int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\ dx = 10\sec^{-1}10x + C\), as \(a = \frac1{10}\).

\(\int \frac{1}{\sqrt{5-x^2}} = \sin^{-1}\frac{x}{\sqrt{5}}+C\), as \(a = \sqrt{5}\).

Most applications of Theorem \(\PageIndex{2}\) are not as straightforward. The next examples show some common integrals that can still be approached with this theorem.

Example \(\PageIndex{14}\): Integrating by substitution: completing the square

Evaluate \( \int\frac{1}{x^2-4x+13}\ dx\).

**Solution**

Initially, this integral seems to have nothing in common with the integrals in Theorem \(\PageIndex{2}\). As it lacks a square root, it almost certainly is not related to arcsine or arcsecant. It is, however, related to the arctangent function.

We see this by *completing the square* in the denominator. We give a brief reminder of the process here.

Start with a quadratic with a leading coefficient of 1. It will have the form of \(x^2 + bx + c\). Take 1/2 of \(b\), square it, and add/subtract it back into the expression. I.e.,

\[\begin{align} x^2+bx+ c &= \underbrace{x^2 + bx + \frac{b^2}4}_{(x+b/2)^2} - \frac{b^2}4 + c\\&= \left(x+\frac b2\right)^2 + c-\frac{b^2}4\end{align}\]

In our example, we take half of \(-4\) and square it, getting \(4\). We add/subtract it into the denominator as follows:

\]\begin{align}\frac{1}{x^2-4x+13} &= \frac{1}{\underbrace{x^2-4x+4}_{(x-2)^2}-4+13}\\ &=\frac{1}{(x-2)^2 + 9}\end{align}\]

We can now integrate this using the arctangent rule. Technically, we need to substitute first with \(u=x-2\), but we can employ Key Idea 10 instead. Thus we have

$$ \int \frac{1}{x^2-4x+13}\ dx = \int \frac{1}{(x-2)^2+9}\ dx = \frac13\tan^{-1}\frac{x-2}{3}+C.$$

Example \(\PageIndex{15}\): Integrals requiring multiple methods

Evaluate \(\int \frac{4-x}{\sqrt{16-x^2}}\ dx\).

**Solution**

This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:

$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\ dx.$$

The first integral is handled using a straightforward application of Theorem \(\PageIndex{2}\); the second integral is handled by substitution, with \(u = 16-x^2\). We handle each separately.

\( \int \frac{4}{\sqrt{16-x^2}}\ dx = 4\sin^{-1}\frac{x}{4} + C.\)

\( \int\frac{x}{\sqrt{16-x^2}}\ dx\): Set \(u = 16-x^2\), so \(du = -2xdx\) and \(xdx = -du/2\). We have

\[\begin{align} \int\frac{x}{\sqrt{16-x^2}}\ dx &= \int\frac{-du/2}{\sqrt{u}}\\ &= -\frac12\int \frac{1}{\sqrt{u}}\ du \\ &= - \sqrt{u} + C\\ &= -\sqrt{16-x^2} + C.\end{align}\]

Combining these together, we have

$$ \int \frac{4-x}{\sqrt{16-x^2}}\ dx = 4\sin^{-1}\frac x4 + \sqrt{16-x^2}+C.$$

Exercise \(\PageIndex{1}\)

Use substitution to find the antiderivative of \(∫3x^2(x^3−3)^2dx.\)

**Hint**-
Let \(u=x^3−3.\)

**Answer**-
\[ ∫3x^2(x^3−3)^2dx=\dfrac{1}{3}(x^3−3)^3+C\]

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

Example \(\PageIndex{2}\): Using Substitution with Alteration

Use substitution to find the antiderivative of \[ ∫z\sqrt{z^2−5}\,dz.\]

**Solution**

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

\[ u=z^2−5\]

\[ du=2z\,dz\]

\[ \dfrac{1}{2}du=\dfrac{1}{2}(2z)\,dz=z\,dz.\]

Write the integral in terms of *u*, but pull the \(\dfrac{1}{2}\) outside the integration symbol:

\[ ∫z(z^2−5)^{1/2}\,dz=\dfrac{1}{2}∫u^{1/2}\,du.\]

Integrate the expression in \(u\):

\(\dfrac{1}{2}∫u^{1/2}\,du=(\dfrac{1}{2})\dfrac{u^{3/2}}{\dfrac{3}{2}}+C\)

\(=(\dfrac{1}{2})(\dfrac{2}{3})u^{3/2}+C\)

\(=\dfrac{1}{3}u^{3/2}+C\)

\(=\dfrac{1}{3}(z^2−5)^{3/2}+C\)

Exercise \(\PageIndex{2}\)

Use substitution to find the antiderivative of \(∫x^2(x^3+5)^9\,dx.\)

**Hint**-
Multiply the du equation by \(\dfrac{1}{3}\).

**Answer**-
\[ \dfrac{(x^3+5)^{10}}{30}+C\]

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 is in terms of the original variable. When we are done, \(u\) should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of \(u\). This technique should become clear in the next example.

Example \(\PageIndex{4}\): Finding an Antiderivative Using u-Substitution

Use substitution to find the antiderivative of \[ ∫\dfrac{x}{\sqrt{x−1}}\,dx.\]

**Solution**

If we let \(u=x−1,\) then \(du=dx\). But this does not account for the *x* in the numerator of the integrand. We need to express x in terms of u. If \(u=x−1\), then \(x=u+1.\) Now we can rewrite the integral in terms of *u*:

\[ ∫\dfrac{x}{\sqrt{x−1}}\,dx=∫\dfrac{u+1}{\sqrt{u}}\,du=∫\sqrt{u}+\dfrac{1}{\sqrt{u}}\,du=∫(u^{1/2}+u^{−1/2})\,du.\]

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

\(∫(u^{1/2}+u^{−1/2})\,du=\dfrac{2}{3}u^{3/2}+2u^{1/2}+C\)

\(=\dfrac{2}{3}(x−1)^{3/2}+2(x−1)^{1/2}+C\)

\(=(x−1)^{1/2}[\dfrac{2}{3}(x−1)+2]+C\)

\(=(x−1)^{1/2}(\dfrac{2}{3}x−\dfrac{2}{3}+\dfrac{6}{3})\)

\(=(x−1)^{1/2}(\dfrac{2}{3}x+\dfrac{4}{3})\)

\(=\dfrac{2}{3}(x−1)^{1/2}(x+2)+C.\)

### Using Substitution with Integrals of Trigonometric Functions

Example \(\PageIndex{4}\): Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral \(∫\dfrac{\sin t}{\cos^3t}\,dt.\)

**Solution**

We know the derivative of \(\cos t\) is \(−\sin t\), so we set \(u=\cos t\). Then \(du=−\sin t\,dt.\)

Substituting into the integral, we have

\[ ∫\dfrac{\sin t}{\cos^3t}\,dt=−∫\dfrac{du}{u^3}.\]

Evaluating the integral, we get

\[ −∫\dfrac{du}{u^3}=−∫u^{−3}\,du=−(−\dfrac{1}{2})u^{−2}+C.\]

Putting the answer back in terms of t, we get

\[ ∫\dfrac{\sin t}{\cos^3t}\,dt=\dfrac{1}{2u^2}+C=\dfrac{1}{2\cos^2t}+C.\]

Exercise \(\PageIndex{4}\)

Use substitution to evaluate the integral \[ ∫\dfrac{\cos t}{\sin^2t}\,dt.\]

**Hint**-
Use the process from Example to solve the problem.

**Answer**-
\(−\dfrac{1}{\sin t}+C\)

Exercise \(\PageIndex{4}\)

Use substitution to evaluate the indefinite integral \[ ∫\cos^3t\sin t\,dt.\]

**Hint**-
Use the process from Example to solve the problem.

**Answer**-
\[ −\dfrac{cos^4t}{4}+C\]

Example \(\PageIndex{5}\):

\[ ∫\sec(t)\,dt.\]

Exercise \(\PageIndex{5}\)

\[ ∫\csc(t)\,dt.\]

**Answer**-
Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page.

### Integrals Involving Logarithmic Functions and involving Exponential Function

A common mistake when dealing with exponential expressions is treating the exponent one the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent one. 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.

Example \(\PageIndex{5}\): Square Root of an Exponential Function

Find the antiderivative of the exponential function \(e^x\sqrt{1+e^x}\).

**Solution**

First rewrite the problem using a rational exponent:

\(∫e^x\sqrt{1+e^x}dx=∫e^x(1+e^x)^{1/2}dx.\)

Using substitution, choose \(u=1+e^x.u=1+e^x\)Then, \(du=e^xdx\). We have (Figure)

\(∫e^x(1+e^x)^{1/2}dx=∫u^{1/2}du.\)

Then

\(∫u^{1/2}du=\dfrac{u^{3/2}}{3/2}+C=\dfrac{2}{3}u^{3/2}+C=\dfrac{2}{3}(1+e^x)^{3/2}+C\)

**Figure \(\PageIndex{1}\):** The graph shows an exponential function times the square root of an exponential function.

Exercise \(\PageIndex{6}\)

Find the antiderivative of \(e^x(3e^x−2)^2\).

**Hint**-
Let \(u=3e^x−2u=3e^x−2.\)

**Answer**-
\(∫e^x(3e^x−2)^2dx=\dfrac{1}{9}(3e^x−2)^3\)

Example \(\PageIndex{7}\): Using Substitution with an Exponential Function

Use substitution to evaluate the indefinite integral \(∫3x^2e^{2x^3}dx.\)

**Solution**

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

\(∫3x^2e^{2x^3}dx=\dfrac{1}{2}∫e^udu\).

Integrate the expression in u and then substitute the original expression in x back into the u integral:

\(\dfrac{1}{2}∫e^udu=\dfrac{1}{2}e^u+C=\dfrac{1}{2}e^2x^3+C.\)

Exercise \(\PageIndex{8}\)

Evaluate the indefinite integral \(∫2x^3ex^4dx\).

**Hint**-
Let \(u=x^4.\)

**Answer**-
\(∫2x^3ex^4dx=\dfrac{1}{2}e^{x^4}\)

Example \(\PageIndex{9}\): Finding an Antiderivative Involving \(\ln x\)

Find the antiderivative of the function \[\dfrac{3}{x−10}.\]

**Solution**

First factor the 3 outside the integral symbol. Then use the \(u^{−1}\) rule. Thus,

\[∫\dfrac{3}{x−10}dx=3∫\dfrac{1}{x−10}dx=3∫\dfrac{du}{u}=3\ln |u|+C=3\ln |x−10|+C,x≠10.\]

See Figure.

**Figure \(\PageIndex{3}\):** The domain of this function is \(x \neq 10.\)

Exercise \(\PageIndex{9}\)

Find the antiderivative of \[\dfrac{1}{x+2}.\]

**Hint**-
Follow the pattern from Example to solve the problem.

**Answer**-
\[\ln |x+2|+C\]

Example \(\PageIndex{10}\): Finding an Antiderivative of a Rational Function

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

**Solution**

This can be rewritten as \(∫(2x^3+3x)(x^4+3x^2)^{−1}dx.\) Use substitution. Let \(u=x^4+3x^2\), then \(du=4x^3+6x.\) Alter du by factoring out the 2. Thus,

\[du=(4x^3+6x)dx=2(2x^3+3x)dx\]

\[\dfrac{1}{2}du=(2x^3+3x)dx.\]

Rewrite the integrand in* u*:

\[∫(2x^3+3x)(x^4+3x^2)^{−1}dx=\dfrac{1}{2}∫u^{−1}du.\]

Then we have

\[\dfrac{1}{2}∫u^{−1}du=\dfrac{1}{2}\ln |u|+C=\dfrac{1}{2}\ln ∣x^4+3x^2∣+C.\]

Example \(\PageIndex{11}\): Finding an Antiderivative of a Logarithmic Function

Find the antiderivative of the log function \[\drac{\ln(2x)}{x}.\)

**Solution**

Let \(u= \ln(2x)\). Then \(du=\dfrac{2}{x} dx\). Now,

\[∫\drac{\ln(2x)}{x}.dx= \int \dfrac{1}{2} {u}du= \dfrac{1}{\2} u^2+C= \dfrac{1}{2}(\ln x)^2+C.\]

## 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 u and du for appropriate expressions in the integrand.
- Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
- Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
- Substitution is often required to put the integrand in the correct form.

## Key Equations

**Substitution with Indefinite Integrals**

\(∫f[g(x)]g′(x)dx=∫f(u)du=F(u)+C=F(g(x))+C\)

**Integrals That Produce Inverse Trigonometric Functions**

\( ∫\dfrac{du}{\sqrt{a^2−u^2}}=sin^{−1}(\dfrac{u}{a})+C\)

\( ∫\dfrac{du}{a^2+u^2}=\dfrac{1}{a}tan^{−1}(\dfrac{u}{a})+C\)

\( ∫\dfrac{du}{u\sqrt{u^2−a^2}}=\dfrac{1}{a}sec^{−1}(\dfrac{u}{a})+C\)

**Contributors**

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/

Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)