Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 6

Integration

Change of Variable: u-identification

As you may suspect, integration is more challenging than differentiation. Fortunately, many people have been integrating before us, and their findings have produced tables of integrals that we can use to facilitate our work. But we must learn to change the variable of the integral given by means of a $u$-identification before we can use them. A table of integrals is provided on page 6 of the textbook. In the discussion that follows, we refer to these integrals by their identification numbers.

Warning: Integration tables facilitate our work, but

• their use is not always the most efficient way of determining an integral.
• they do not include all of the possible integrals we may encounter.

So, you must learn as many techniques of integration as possible.

If ${\displaystyle u=f\left(x\right)\text{,}}$ then its differential is $f^{\prime }\left(x\right)dx$. Thus, by identifying the integrand function $f(x)$ with ${\displaystyle u\text{,}}$ we find that the integral

${\displaystyle \int f{\left(x\right)}^r{f}^{\prime }\left(x\right)dx={\displaystyle \frac{{\left(f\left(x\right)\right)}^{r+1}}{r+1}}}$

is the integral

${\displaystyle \int u^{r}du={\displaystyle \frac{u^{r+1}}{r+1}}\text{.}}$

This is Integral ${\displaystyle 2}$ from the table of integrals on page ${\displaystyle 6}$ of the textbook.

Example 6.17. In the integral

${\displaystyle \int {\left(x+1\right)}^{2}dx}$

the $u$ identification is ${\displaystyle u=x+1\text{,}}$ ${\displaystyle r=2\text{,}}$ and ${\displaystyle du=1dx\text{.}}$ Then,

${\displaystyle \int \left(x+1\right)dx=\int u^{2}du={\displaystyle \frac{u^3}{3}}={\displaystyle \frac{{\left(x+1\right)}^3}{3}}+C\text{.}}$

Comparing this with the method we use in part A, we see that the $u$-identification is not more efficient in this case, because the integrand function is relatively simple.

Example 6.18. In the integral ${\displaystyle \int x{\left(x^2+1\right)}^{3}dx}$ of part C, we identify ${\displaystyle u=x^2+1\text{,}}$ ${\displaystyle r=3\text{.}}$

Hence, ${\displaystyle du=2xdx\text{,}}$ and ${\displaystyle {\displaystyle \frac{1}{2}}du=xdx\text{.}}$

Thus,

${\displaystyle \int x{(x^2+1)}^{3}dx}={\displaystyle \int u^3{\displaystyle \frac{1}{2}}du}={\displaystyle \frac{1}{2}}{\displaystyle \int u^{3}du}={\displaystyle \frac{1}{2}}{\displaystyle \frac{u^4}{4}}={\displaystyle \frac{{(x^2+1)}^4}{8}}+C\text{.}$

Example 6.19. In the integral ${\displaystyle \int {\cos }^3\left(4x\right)\sin \left(4x\right)dx\text{,}}$ we identify ${\displaystyle u=\cos \left(4x\right)\text{,}}$ ${\displaystyle r=3\text{.}}$

Then ${\displaystyle du=-4\sin \left(4x\right)dx}$, and ${\displaystyle -{\displaystyle \frac{1}{4}}du=\sin \left(4x\right)dx}$.

Thus,

$\begin{array}{rl}{\displaystyle \int {\cos }^3(4x)\sin (4x)dx}& ={\displaystyle \int u^3-{\displaystyle \frac{1}{4}}du}\\ & =\left(-{\displaystyle \frac{1}{4}}\right){\displaystyle \int u^{3}du}\\ & =-{\displaystyle \frac{u^4}{4(4)}}\\ & =-{\displaystyle \frac{{\cos }^4(4x)}{16}}\text{.}\end{array}$

Compare this result with Example 6.15.

Observe that, once we identify ${\displaystyle u\text{,}}$ we obtain the differential ${\displaystyle du\text{,}}$ and in order to apply the corresponding integral from the table, we must rewrite the given integral only in terms of ${\displaystyle u\text{.}}$ However, the answer must be given in terms of the original variable of the integrand function, not in terms of ${\displaystyle u\text{.}}$

Example 6.20. If we compare the integral ${\displaystyle \int x\sin \left(1-2x^2\right)dx}$ with the integrals in the table of integrals, we see that Integral ${\displaystyle 6}$ is the closest, if we identify ${\displaystyle u=1-2x^2\text{,}}$ because ${\displaystyle du=-4xdx\text{.}}$

So, we have ${\displaystyle {\displaystyle \frac{1}{-4}}du=xdx}$; therefore,

${\displaystyle \begin{array}{ll}{\displaystyle \int x\sin \left(1-2x^2\right)dx}\hfill & ={\displaystyle \int \sin u}\left({\displaystyle \frac{1}{-4}}\right)du=-{\displaystyle \frac{1}{4}}{\displaystyle \int \sin udu}\hfill \\ \hfill & =-\left({\displaystyle \frac{1}{4}}\right)\left(-\cos u\right)={\displaystyle \frac{\cos \left(1-2x^2\right)}{4}}+C\text{.}\hfill \end{array}}$

To be able to use integration tables efficiently, you must be aware that the differential

${\displaystyle du=f^{\prime }\left(x\right)dx}$

is the derivative of a function. So, you must identify a function and its derivative (or an expression that is close to its derivative, up to a constant), in the integrand function. Your ability to do so is the key in applying integration tables properly. Observe the general forms of the integrals in the table, because you will always start with a general form in terms of a variable other than ${\displaystyle u\text{.}}$ For instance,

${\displaystyle \begin{array}{llll}\text{Integral 7}\text{.}\hfill & {\displaystyle \int \cos udu}\hfill & \text{the general form is}\hfill & {\displaystyle \int \cos \left(f\left(x\right)\right)f^{\prime }\left(x\right)dx\text{.}}\hfill \end{array}}$

${\displaystyle \begin{array}{llll}\text{Integral 8}\text{.}\hfill & {\displaystyle \int {\text{sec}}^{2}udu}\hfill & \text{the general form is}\hfill & {\displaystyle \int {\sec }^2\left(f\left(x\right)\right)f^{\prime }\left(x\right)dx\text{.}}\hfill \end{array}}$

The integral

${\displaystyle \int \cos \left(x^3\right)xdx}$

does not fit the form of Integral 7, because we have ${\displaystyle x^3}$ and $x$ in the integrand function, and neither one of them is, or almost is, the derivative of the other. Hence, we do not have $f(x)$ and ${\displaystyle f^{\prime }\left(x\right)}$ in the integrand function.

However, the integral

${\displaystyle \int \cos \left(x^3\right)x^{2}dx}$

does fit the form of Integral 7, because we have ${\displaystyle x^3}$ and ${\displaystyle x^2\text{,}}$ the latter being almost the derivative of the former. Hence, the $u$-identification that we use is ${\displaystyle u=x^3}$ and ${\displaystyle du=3x^{2}dx}$:

${\displaystyle \int \cos \left(x^3\right)x^{2}dx={\displaystyle \frac{1}{3}}\int \cos udu={\displaystyle \frac{\sin \left(x^3\right)}{3}}+C\text{.}}$

Note: Whenever you use an integral from the table of integrals provided in the textbook, you must identify it by number, and give the explicit identification of $\color{#384877}{u}$ and the differential $\color{#384877}{du}$.