Skills Review for Calculus of Vector-Valued Functions

Learning Outcomes

Apply basic derivative rules
Use the product rule for finding the derivative of a product of functions
Use the quotient rule for finding the derivative of a quotient of functions
Apply the chain rule together with the power and product rule
Evaluate indefinite integrals

In the Calculus of Vector-Valued Functions section, we will learn how to differentiate and integrate vector-valued functions. Here we will review various derivative rules and integration techniques.

Basic Derivative Rules

We first apply the limit definition of the derivative to find the derivative of the constant function, $f(x)=c$ . For this function, both $f(x)=c$ and $f(x+h)=c$ , so we obtain the following result:

\begin{array}{ll} f^{'} (x) & = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{h \to 0} \frac{c - c}{h} \\ = lim_{h \to 0} \frac{0}{h} \\ = lim_{h \to 0} 0 = 0 \end{array}

$\begin{array}{ll}f^{\prime}(x) & =\underset{h\to 0}{\lim}\dfrac{f(x+h)-f(x)}{h} \\ & =\underset{h\to 0}{\lim}\dfrac{c-c}{h} \\ & =\underset{h\to 0}{\lim}\dfrac{0}{h} \\ & =\underset{h\to 0}{\lim}0=0 \end{array}$

The rule for differentiating constant functions is called the constant rule. It states that the derivative of a constant function is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is 0. We restate this rule in the following theorem.

The Constant Rule

Let $c$ be a constant.

If $f(x)=c$ , then $f^{\prime}(c)=0$

Alternatively, we may express this rule as

$\dfrac{d}{dx}(c)=0$

Example: Applying the Constant Rule

Find the derivative of $f(x)=8$ .

Show Solution

This is just a one-step application of the rule:

f^{'} (8) = 0

$f^{\prime}(8)=0$ .

Try It

Find the derivative of $g(x)=-3$ .

Show Solution

We have shown that

\frac{d}{d x} (x^{2}) = 2 x

$\dfrac{d}{dx}\left(x^2\right)=2x$ and

\frac{d}{d x} (x^{\frac{1}{2}}) = \frac{1}{2} x^{- \frac{1}{2}}

$\dfrac{d}{dx}\left(x^{\frac{1}{2}}\right)=\dfrac{1}{2}x^{−\frac{1}{2}}$

At this point, you might see a pattern beginning to develop for derivatives of the form $\frac{d}{dx}(x^n)$ . We continue our examination of derivative formulas by differentiating power functions of the form $f(x)=x^n$ where $n$ is a positive integer. We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating and proving the general rule for derivatives of functions of this form, we take a look at a specific case, $\frac{d}{dx}(x^3)$ .

As we shall see, the procedure for finding the derivative of the general form $f(x)=x^n$ is very similar. Although it is often unwise to draw general conclusions from specific examples, we note that when we differentiate $f(x)=x^3$ , the power on $x$ becomes the coefficient of $x^2$ in the derivative and the power on $x$ in the derivative decreases by 1. The following theorem states that this power rule holds for all positive integer powers of $x$ . We will eventually extend this result to negative integer powers. Later, we will see that this rule may also be extended first to rational powers of $x$ and then to arbitrary powers of $x$ . Be aware, however, that this rule does not apply to functions in which a constant is raised to a variable power, such as $f(x)=3^x$ .

The Power Rule

Let $n$ be a positive integer. If $f(x)=x^n$ , then

f^{'} (x) = n x^{n - 1}

$f^{\prime}(x)=nx^{n-1}$

Alternatively, we may express this rule as

\frac{d}{d x} (x^{n}) = n x^{n - 1}

$\dfrac{d}{dx}(x^n)=nx^{n-1}$

Example: Applying the Power Rule

Find the derivative of the function $f(x)=x^{10}$ by applying the power rule.

Show Solution

Using the power rule with $n=10$ , we obtain

f^{'} (x) = 10 x^{10 - 1} = 10 x^{9}

$f^{\prime}(x)=10x^{10-1}=10x^9$ .

Try It

Find the derivative of $f(x)=x^7$ .

Hint

Use the power rule with $n=7$ .

Show Solution

$f^{\prime}(x)=7x^6$

We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions. Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract, or multiply by a constant. These rules are summarized in the following theorem.

Sum, Difference, and Constant Multiple Rules

Let $f(x)$ and $g(x)$ be differentiable functions and $k$ be a constant. Then each of the following equations holds.

Sum Rule: The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g$ .

\frac{d}{d x} (f (x) + g (x)) = \frac{d}{d x} (f (x)) + \frac{d}{d x} (g (x))

$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$ ;

that is,

for

j (x) = f (x) + g (x), j^{'} (x) = f^{'} (x) + g^{'} (x)

$j(x)=f(x)+g(x), \, j^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)$

Difference Rule: The derivative of the difference of a function $f$ and a function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$ .

\frac{d}{d x} (f (x) - g (x)) = \frac{d}{d x} (f (x)) - \frac{d}{d x} (g (x))

$\frac{d}{dx}(f(x)-g(x))=\frac{d}{dx}(f(x))-\frac{d}{dx}(g(x))$ ;

that is,

for

j (x) = f (x) - g (x), j^{'} (x) = f^{'} (x) - g^{'} (x)

$j(x)=f(x)-g(x), \, j^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)$

Constant Multiple Rule: The derivative of a constant $k$ multiplied by a function $f$ is the same as the constant multiplied by the derivative:

\frac{d}{d x} (k f (x)) = k \frac{d}{d x} (f (x))

$\frac{d}{dx}(kf(x))=k\frac{d}{dx}(f(x))$ ;

that is,

for

j (x) = k f (x), j^{'} (x) = k f^{'} (x)

$j(x)=kf(x), \, j^{\prime}(x)=kf^{\prime}(x)$

Example: Applying Basic Derivative Rules

Find the derivative of $f(x)=2x^5+7$ .

Show Solution

We begin by applying the rule for differentiating the sum of two functions, followed by the rules for differentiating constant multiples of functions and the rule for differentiating powers. To better understand the sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution:

\begin{array}{lllll} f^{'} (x) & = \frac{d}{d x} (2 x^{5} + 7) \\ = \frac{d}{d x} (2 x^{5}) + \frac{d}{d x} (7) & Apply the sum rule. \\ = 2 \frac{d}{d x} (x^{5}) + \frac{d}{d x} (7) & Apply the constant multiple rule. \\ = 2 (5 x^{4}) + 0 & Apply the power rule and the constant rule. \\ = 10 x^{4} . & Simplify. \end{array}

$\begin{array}{lllll}f^{\prime}(x) & =\frac{d}{dx}(2x^5+7) & & & \\ & =\frac{d}{dx}(2x^5)+\frac{d}{dx}(7) & & & \text{Apply the sum rule.} \\ & =2\frac{d}{dx}(x^5)+\frac{d}{dx}(7) & & & \text{Apply the constant multiple rule.} \\ & =2(5x^4)+0 & & & \text{Apply the power rule and the constant rule.} \\ & =10x^4. & & & \text{Simplify.} \end{array}$

Try It

Find the derivative of $f(x)=2x^3-6x^2+3$ .

Hint

Show Solution

$f^{\prime}(x)=6x^2-12x$ .

The Product Rule

Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function $f(x)=x^2$ , whose derivative is $f^{\prime}(x)=2x$ and not $\frac{d}{dx}(x)\cdot \frac{d}{dx}(x)=1\cdot 1=1$ .

Product Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

\frac{d}{d x} (f (x) g (x)) = \frac{d}{d x} (f (x)) \cdot g (x) + \frac{d}{d x} (g (x)) \cdot f (x)

$\frac{d}{dx}(f(x)g(x))=\frac{d}{dx}(f(x))\cdot g(x)+\frac{d}{dx}(g(x))\cdot f(x)$

That is,

j (x) = f (x) g (x)

$j(x)=f(x)g(x)$ then

j^{'} (x) = f^{'} (x) g (x) + g^{'} (x) f (x)

$j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)$

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Example: Applying the Product Rule to Binomials

For $j(x)=(x^2+2)(3x^3-5x)$ , find $j^{\prime}(x)$ by applying the product rule.

Show Solution

If we set $f(x)=x^2+2$ and $g(x)=3x^3-5x$ , then $f^{\prime}(x)=2x$ and $g^{\prime}(x)=9x^2-5$ . Thus,

j^{'} (x) = f^{'} (x) g (x) + g^{'} (x) f (x) = (2 x) (3 x^{3} - 5 x) + (9 x^{2} - 5) (x^{2} + 2)

$j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2)$ .

Simplifying, we have

j^{'} (x) = 15 x^{4} + 3 x^{2} - 10

$j^{\prime}(x)=15x^4+3x^2-10$ .

Try It

Use the product rule to obtain the derivative of $j(x)=2x^5(4x^2+x)$ .

Hint

Set $f(x)=2x^5$ and $g(x)=4x^2+x$ and use the preceding example as a guide.

Show Solution

$j^{\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5$ .

The Quotient Rule

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

\frac{d}{d x} (x^{2}) = 2 x

$\frac{d}{dx}(x^2)=2x$ , which is not the same as

\frac{\frac{d}{d x} (x^{3})}{\frac{d}{d x} (x)} = \frac{3 x^{2}}{1} = 3 x^{2}

$\dfrac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(x)} =\dfrac{3x^2}{1}=3x^2$

The Quotient Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

\frac{d}{d x} (\frac{f (x)}{g (x)}) = \frac{\frac{d}{d x} (f (x)) \cdot g (x) - \frac{d}{d x} (g (x)) \cdot f (x)}{(g (x))^{2}}

$\frac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\frac{d}{dx}(f(x))\cdot g(x)-\dfrac{d}{dx}(g(x))\cdot f(x)}{(g(x))^2}$

That is,

j (x) = \frac{f (x)}{g (x)}

$j(x)=\dfrac{f(x)}{g(x)}$ , then

j^{'} (x) = \frac{f^{'} (x) g (x) - g^{'} (x) f (x)}{(g (x))^{2}}

$j^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}$

Example: Applying the Quotient Rule

Use the quotient rule to find the derivative of $k(x)=\dfrac{5x^2}{4x+3}$

Show Solution

Let $f(x)=5x^2$ and $g(x)=4x+3$ . Thus, $f^{\prime}(x)=10x$ and $g^{\prime}(x)=4$ . Substituting into the quotient rule, we have

k^{'} (x) = \frac{f^{'} (x) g (x) - g^{'} (x) f (x)}{(g (x))^{2}} = \frac{10 x (4 x + 3) - 4 (5 x^{2})}{(4 x + 3)^{2}}

$k^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}=\dfrac{10x(4x+3)-4(5x^2)}{(4x+3)^2}$ .

Simplifying, we obtain

k^{'} (x) = \frac{20 x^{2} + 30 x}{(4 x + 3)^{2}}

$k^{\prime}(x)=\dfrac{20x^2+30x}{(4x+3)^2}$ .

Try It

Find the derivative of $h(x)=\dfrac{3x+1}{4x-3}$

Hint

Apply the quotient rule with $f(x)=3x+1$ and $g(x)=4x-3$ .

Show Solution

$k^{\prime}(x)=-\dfrac{13}{(4x-3)^2}$ .

The Chain Rule

Let $f$ and $g$ be functions. For all $x$ in the domain of $g$ for which $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , the derivative of the composite function

h (x) = (f \circ g) (x) = f (g (x))

$h(x)=(f\circ g)(x)=f(g(x))$

is given by

h^{'} (x) = f^{'} (g (x)) g^{'} (x)

$h^{\prime}(x)=f^{\prime}(g(x))g^{\prime}(x)$

Alternatively, if $y$ is a function of $u$ , and $u$ is a function of $x$ , then

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}$

Note that we often need to use the chain rule with other rules. For example, to find derivatives of functions of the form $h(x)=(g(x))^n$ , we need to use the chain rule combined with the power rule. To do so, we can think of $h(x)=(g(x))^n$ as $f(g(x))$ where $f(x)=x^n$ . Then $f^{\prime}(x)=nx^{n-1}$ . Thus, $f^{\prime}(g(x))=n(g(x))^{n-1}$ . This leads us to the derivative of a power function using the chain rule,

h^{'} (x) = n (g (x))^{n - 1} g^{'} (x)

$h^{\prime}(x)=n(g(x))^{n-1}g^{\prime}(x)$

Power Rule for Composition of Functions

For all values of $x$ for which the derivative is defined, if

h (x) = (g (x))^{n}

$h(x)=(g(x))^n$

Then

h^{'} (x) = n (g (x))^{n - 1} g^{'} (x)

$h^{\prime}(x)=n(g(x))^{n-1}g^{\prime}(x)$

Example: Using the Chain and Power Rules

Find the derivative of $h(x)=\dfrac{1}{(3x^2+1)^2}$

Show Solution

First, rewrite $h(x)=\frac{1}{(3x^2+1)^2}=(3x^2+1)^{-2}$ .

Applying the power rule with $g(x)=3x^2+1$ , we have

h^{'} (x) = - 2 (3 x^{2} + 1)^{- 3} (6 x)

$h^{\prime}(x)=-2(3x^2+1)^{-3}(6x)$ .

Rewriting back to the original form gives us

h^{'} (x) = \frac{- 12 x}{(3 x^{2} + 1)^{3}}

$h^{\prime}(x)=\frac{-12x}{(3x^2+1)^3}$ .

Try It

Find the derivative of $h(x)=(2x^3+2x-1)^4$

Show Solution

$h^{\prime}(x)=4(2x^3+2x-1)^3(6x^{2}+2)=8(3x^{2}+1)(2x^3+2x-1)^3$

Hint

Use the previous example with $g(x)=2x^3+2x-1$

Example: Using the Chain and Power Rules with a Trigonometric Function

Find the derivative of $h(x)=\sin^3 x$

Show Solution

First recall that $\sin^3 x=(\sin x)^3$ , so we can rewrite $h(x)= \sin^3 x$ as $h(x)=(\sin x)^3$ .

Applying the power rule with $g(x)= \sin x$ , we obtain

h^{'} (x) = 3 (\sin x)^{2} \cos x = 3 \sin^{2} x \cos x

$h^{\prime}(x)=3(\sin x)^2 \cos x=3 \sin^2 x \cos x$ .

Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.

Example: Using the Chain Rule on a Cosine Function

Find the derivative of $h(x)= \cos (5x^2)$ .

Show Solution

Let

g (x) = 5 x^{2}

$g(x)=5x^2$ . Then

g^{'} (x) = 10 x

$g^{\prime}(x)=10x$ .
Using the result from the previous example,

\begin{array}{ll} h^{'} (x) & = - \sin (5 x^{2}) \cdot 10 x \\ = - 10 x \sin (5 x^{2}) \end{array}

$\begin{array}{ll}h^{\prime}(x) & =-\sin (5x^2) \cdot 10x \\ & =-10x \sin (5x^2) \end{array}$

Example: Using the Chain Rule on Another Trigonometric Function

Find the derivative of $h(x)= \sec (4x^5+2x)$ .

Show Solution

Apply the chain rule to $h(x)= \sec (g(x))$ to obtain

h^{'} (x) = \sec (g (x)) \tan (g (x)) g^{'} (x)

$h^{\prime}(x)= \sec (g(x)) \tan (g(x))g^{\prime}(x)$ .

In this problem, $g(x)=4x^5+2x$ , so we have $g^{\prime}(x)=20x^4+2$ . Therefore, we obtain

\begin{array}{ll} h^{'} (x) & = \sec (4 x^{5} + 2 x) \tan (4 x^{5} + 2 x) (20 x^{4} + 2) \\ = (20 x^{4} + 2) \sec (4 x^{5} + 2 x) \tan (4 x^{5} + 2 x) \end{array}

$\begin{array}{ll}h^{\prime}(x) & = \sec (4x^5+2x) \tan (4x^5+2x)(20x^4+2) \\ & =(20x^4+2) \sec (4x^5+2x) \tan (4x^5+2x) \end{array}$

Try It

Find the derivative of $h(x)= \sin (7x+2)$ .

Hint

Apply the chain rule to $h(x)= \sin g(x)$ first and then use $g(x)=7x+2$ .

Show Solution

$h^{\prime}(x)=7 \cos (7x+2)$

We now provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions.

Using the Chain Rule with Trigonometric Functions

For all values of $x$ for which the derivative is defined,

$\begin{array}{llll}\frac{d}{dx}(\sin (g(x)))= \cos (g(x))g^{\prime}(x) & & & \frac{d}{dx} \sin u= \cos u\frac{du}{dx} \\ \frac{d}{dx}(\cos (g(x)))=−\sin (g(x))g^{\prime}(x) & & & \frac{d}{dx} \cos u=−\sin u\frac{du}{dx} \\ \frac{d}{dx}(\tan (g(x)))= \sec^2 (g(x))g^{\prime}(x) & & & \frac{d}{dx} \tan u=\sec^2 u\frac{du}{dx} \\ \frac{d}{dx}(\cot (g(x)))=−\csc^2 (g(x))g^{\prime}(x) & & & \frac{d}{dx} \cot u=−\csc^2 u\frac{du}{dx} \\ \frac{d}{dx}(\sec (g(x)))= \sec (g(x)) \tan (g(x))g^{\prime}(x) & & & \frac{d}{dx} \sec u= \sec u \tan u\frac{du}{dx} \\ \frac{d}{dx}(\csc (g(x)))=−\csc (g(x)) \cot (g(x))g^{\prime}(x) & & & \frac{d}{dx} \csc u=−\csc u \cot u\frac{du}{dx} \end{array}$

Indefinite Integrals

Power Rule for Integrals

For $n \ne −1$ ,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

$\displaystyle\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$

Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in Appendix B: Table of Derivatives.

Integration Formulas
Differentiation Formula	Indefinite Integral
$\frac{d}{dx}(k)=0$	$\displaystyle\int kdx=\displaystyle\int kx^0 dx=kx+C$
$\frac{d}{dx}(x^n)=nx^{n-1}$	$\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne −1$
$\frac{d}{dx}(\ln \|x\|)=\frac{1}{x}$	$\displaystyle\int \frac{1}{x}dx=\ln \|x\|+C$
$\frac{d}{dx}(e^x)=e^x$	$\displaystyle\int e^x dx=e^x+C$
$\frac{d}{dx}(\sin x)= \cos x$	$\displaystyle\int \cos x dx= \sin x+C$
$\frac{d}{dx}(\cos x)=− \sin x$	$\displaystyle\int \sin x dx=− \cos x+C$
$\frac{d}{dx}(\tan x)= \sec^2 x$	$\displaystyle\int \sec^2 x dx= \tan x+C$
$\frac{d}{dx}(\csc x)=−\csc x \cot x$	$\displaystyle\int \csc x \cot x dx=−\csc x+C$
$\frac{d}{dx}(\sec x)= \sec x \tan x$	$\displaystyle\int \sec x \tan x dx= \sec x+C$
$\frac{d}{dx}(\cot x)=−\csc^2 x$	$\displaystyle\int \csc^2 x dx=−\cot x+C$
$\frac{d}{dx}( \sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}$	$\displaystyle\int \frac{1}{\sqrt{1-x^2}} dx= \sin^{-1} x+C$
$\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}$	$\displaystyle\int \frac{1}{1+x^2} dx= \tan^{-1} x+C$
$\frac{d}{dx}(\sec^{-1} \|x\|)=\frac{1}{x\sqrt{x^2-1}}$	$\displaystyle\int \frac{1}{x\sqrt{x^2-1}} dx= \sec^{-1} \|x\|+C$

Example: Evaluating Indefinite Integrals

Evaluate each of the following indefinite integrals:

$\displaystyle\int (5x^3-7x^2+3x+4) dx$
$\displaystyle\int \frac{x^2+4\sqrt[3]{x}}{x} dx$
$\displaystyle\int \frac{4}{1+x^2} dx$
$\displaystyle\int \tan x \cos x dx$

Show Solution

Using the properties of indefinite integrals, we can integrate each of the four terms in the integrand separately. We obtain
$\displaystyle\int (5x^3-7x^2+3x+4) dx=\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx$

From the Constant Multiples property of indefinite integrals, each coefficient can be written in front of the integral sign, which gives

$\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx=5\displaystyle\int x^3 dx-7\displaystyle\int x^2 dx+3\displaystyle\int x dx+4\displaystyle\int 1 dx$

Using the power rule for integrals, we conclude that

$\displaystyle\int (5x^3-7x^2+3x+4) dx=\frac{5}{4}x^4-\frac{7}{3}x^3+\frac{3}{2}x^2+4x+C$
Rewrite the integrand as
$\frac{x^2+4\sqrt[3]{x}}{x}=\frac{x^2}{x}+\frac{4\sqrt[3]{x}}{x}$

Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have

$\begin{array}{ll} \displaystyle\int (x+\frac{4}{x^{2/3}}) dx & =\displaystyle\int x dx+4\displaystyle\int x^{-2/3} dx \\ & =\frac{1}{2}x^2+4\frac{1}{(\frac{-2}{3})+1}x^{(-2/3)+1}+C \\ & =\frac{1}{2}x^2+12x^{1/3}+C \end{array}$
Using the properties of indefinite integrals, write the integral as
$4\displaystyle\int \frac{1}{1+x^2} dx$ .

Then, use the fact that $\tan^{-1} (x)$ is an antiderivative of $\frac{1}{1+x^2}$ to conclude that

$\displaystyle\int \frac{4}{1+x^2} dx=4 \tan^{-1} (x)+C$
Rewrite the integrand as
$\tan x \cos x=\frac{ \sin x}{ \cos x} \cos x= \sin x$ .

Therefore,

$\displaystyle\int \tan x \cos x dx=\displaystyle\int \sin x dx=− \cos x+C$

Try It

Evaluate $\displaystyle\int (4x^3-5x^2+x-7) dx$

Hint

Show Solution

$x^4-\frac{5}{3}x^3+\frac{1}{2}x^2-7x+C$

Vector-Valued Functions: Prerequisite Material