Antiderivatives

Primitive Functions Given a function $f(x)$

defined on some interval $I$

. Then a function $F(x)$

safisfies

$\displaystyle F^{\prime}(x) = f(x)$

for all $x \in I$ is called primitive function of $f(x)$

Understanding
$\displaystyle{(\frac{x^{3}}{3})^{\prime} = x^{2}}$ . Then a function $\displaystyle{\frac{x^{3}}{3}}$ is a primitive function of $\displaystyle{f(x) = x^{2}}$ . Similarly, since $\displaystyle{(\frac{x^{3}}{3} + 1)^{\prime} = x^2}$ , $\displaystyle{\frac{x^{3}}{3} + 1}$ is a primitive function of $\displaystyle{f(x) = x^2}$ .

Antiderivatives

Theorem 3..1 Let $F(x)$

be a primitive function of $f(x)$

. Then every primitive function of $f(x)$

is given by $F(x) + c$

, where

is an arbitrary constant.

Proof Let $G(x)$ be a primitive function of $f(x)$ . Then $G^{\prime}(x) = f(x)$ and since $F'(x) = f(x)$ , $G^{\prime}(x) = F^{\prime}(x)$ . Now let $H(x) = G(x) - F(x)$ . Then $H'(x) = 0$ . This means that $c = H(x) = G(x) - F(x)$ , where $c$ is constant. Thus $G(x) = F(x) + c$ . Note that $(F(x) + c)^{\prime} = F^{\prime}(x)$ $\blacksquare$

Understanding
Let and be primitive functions of . Then . This means that the slope of the tangent line to and the slope of the tangent line to is the same for all . Thus the graph of and are parallet. Thus, the difference of and is constant.

Antiderivatives Every primitive function of $f(x)$ is called a antiderivative and denoted by $\int f(x)dx$ .

Antiderivative Suppoe $F(x)$ is a primitive function of $f(x)$ . Then $\int f(x) dx = F(x) + c$ . We call this $f(x)$ integrand.

The process of finding an antiderivative of $f(x)$ is calle antidifferentiation or integration.

Example 3..1 Integrate the following function

$\displaystyle \int (x^{-\frac{3}{2}} + \cos{x}) dx$

SOLUTION Note that $\displaystyle{(-2x^{-\frac{1}{2}})' = (-2)(-\frac{1}{2})x^{-\frac{3}{2}} = x^{-\frac{3}{2}}, (\sin{x})' = \cos{x}}$ . Then since the derivative of a sum is the sum of the derivatives,

$\displaystyle (-2x^{-\frac{1}{2}} + \sin{x})' = x^{-\frac{3}{2}} + \cos{x}.$

$\displaystyle{\int (x^{-\frac{3}{2}} + \cos{x})\:dx = -2x^{-\frac{1}{2}} + \sin{x} + c \ensuremath{ \blacksquare}}$

Exercise 3..1 Integrate the following function

$\displaystyle \int (\frac{\sin{x}}{\cos^{2}{x}} + \frac{1}{x}) dx$

SOLUTION Note that

$\displaystyle (\frac{1}{\cos{x}})' = \frac{\sin{x}}{\cos^{2}{x}}, (\log{x})' = \frac{1}{x}.$

Then since the derivative of a sum is the sum of the derivatives,

$\displaystyle (\frac{1}{\cos{x}} + \log{x})' = \frac{\sin{x}}{\cos^{2}{x}} + \frac{1}{x}.$

Thus, $\displaystyle{\int (\frac{\sin{x}}{\cos^{2}{x}} + \frac{1}{x})\:dx = \frac{\sin{x}}{\cos^{2}{x}} + \log{x} + c \ensuremath{ \blacksquare}}$

Check
Find satisfying $F'(x) = \frac{\sin{x}}{\cos^{2}{x}} + \frac{1}{x}$ . $(\frac{1}{\cos{x}})' = \frac{-(\cos{x})'}{\cos^{2}{x}} = \frac{\sin{x}}{\cos^{2}{x}}$ .

In Example3.1 and Exercise3.1, we are somehow able to find an antiderivatives. To integrate a function, it is better to use a integration founula. Integratio formula 1. through 5. , Simply memorize. Integration formila 6. through 12. can be derived by the method of u-substitution. Trig Fcts Express $\frac{1}{\cos{x}}$ as $\sec{x}$ . Express $\frac{1}{\tan{x}}$ as $\cot{x}$ . Integration Formulas

Theorem 3..2 $\displaystyle{1. \int x^{\alpha} dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c \hskip 0.3cm {\rm where} \alpha \neq -1}$

$\displaystyle{2. \int \frac{1}{x} dx = \log{\vert x\vert} + c}$

$\displaystyle{3. \int e^{x} dx = e^{x} + c , \int a^x dx = \frac{a^x}{\log{a}} + c}$ where $a > 0$

$\displaystyle{4. \int \sin{x}dx = -\cos{x} + c}$

$\displaystyle{5. \int \cos{x} dx = \sin{x} + c }$

$\displaystyle{6. \int \tan{x} dx = \log{\vert\sec{x}\vert} + c} = -\log\vert\cos{x}\vert + c$

$\displaystyle{7. \int \frac{1}{\cos^{2}{x}} = \tan{x} + c}$

$\displaystyle{8. \int \frac{1}{\sin^{2}{x}} = -\cot{x} +c}$

$\displaystyle{9. \int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a}\log{\vert\frac{x-a}{x+a}\vert} + c}$

$\displaystyle{10. \int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}{\frac{x}{a}} + c}$ where $a \neq 0$

$\displaystyle{11. \int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}{\frac{x}{a}} + c }$ where $\vert x\vert < a$

$\displaystyle{12. \int \frac{dx}{\sqrt{x^{2} + a}} = \log{\vert x + \sqrt{x^{2} + a}\vert} + c}$ where $x^2 + a > 0$

Theorem3.2-10.
The derivative of $\tan^{-1}{x}$ is already derived in Exercise2.7 and is $(\tan^{-1}{x})' = \frac{1}{1+x^2}$ .

NOTE Differentiate the right-hand side to get the integrand.
1. $\displaystyle{(\frac{x^{\alpha + 1}}{\alpha + 1} + c)^{\prime} = \frac{\alpha + 1}{\alpha + 1}x^{\alpha + 1 - 1} + c' = x^{\alpha}}$ . Thus, $\displaystyle{\int x^{\alpha} dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c}$
10.

$\displaystyle \left(\frac{1}{a}\tan^{-1}{\frac{x}{a}}\right)'$

$\displaystyle =$

$\displaystyle \frac{1}{a} \frac{1}{1 + \left(\frac{x}{a}\right)^2}\cdot \left(\... ...ght)' = \frac{1}{a^2}\cdot \frac{1}{1 + \frac{x^2}{a^2}} = \frac{1}{a^2 + x^2}.$

Rules of Integration

Theorem 3..3 Suppose that integral of $f(x)$

and

exist. Then

$\displaystyle{1. \int\{f(x) \pm g(x)\}dx = \int f(x)dx \pm \int g(x)dx}$

$\displaystyle{2. \int kf(x)dx = k\int f(x)dx} {\rm where}$ k ${\rm is constant}$

$\displaystyle{3. \int \frac{f'(x)}{f(x)}dx = \log\vert f(x)\vert + c}$

NOTE

$\displaystyle 1. \frac{d}{dx}(\int f(x)dx \pm \int g(x)dx)$	$\displaystyle =$	$\displaystyle \frac{d}{dx}\int f(x)dx \pm \frac{d}{dx}\int g(x)dx.$
	$\displaystyle =$	$\displaystyle f(x) \pm g(x).$

Thus, $\displaystyle{\int f(x)dx \pm \int g(x)dx = \int f(x)dx \pm \int g(x)dx}$
2. $\displaystyle{\frac{d}{dx}(k\int f(x)dx) = k \frac{d}{dx}(\int f(x)dx ) = k f(x)}$
3.

$\displaystyle \frac{d(\log\vert f(x)\vert)}{dx} = \frac{d(\log\vert u\vert)}{du}\frac{du}{dx} = \frac{1}{u}f'(x) = \frac{f'(x)}{f(x)}.$

Thus, ,

$\displaystyle \int \frac{f'(x)}{f(x)}\:dx = \log\vert f(x)\vert + c.$

Rules of Integration
1. The integral of a sum is the sum of the integrals. 2. The integral of a constant multiple is the constant multiple of the integral. 3. The integral of a function whose numerator is the derivative of the denominator is a log of the absolute value of the denominator.

Example 3..2 Integrate the following functions.
1. $\displaystyle{3\sin{x} + x^2}$ 2. $\displaystyle{\frac{1}{3 + x^2}}$ 3. $\displaystyle{\frac{1}{\sqrt{x^2 + 5}}}$

In this problem, the rules of the integral of a sum and the interal of a constant multiple can be used. SOLUTION 1.

$\displaystyle \int(3\sin{x} + x^2)dx$	$\displaystyle \underbrace{=}_{{\rm Theorem}3.3-1}$	$\displaystyle \int 3\sin{x} dx + \int x^2 dx$
	$\displaystyle \underbrace{=}_{{\rm Theorem}3.3-2}$	$\displaystyle 3\int \sin{x} dx + \int x^2 dx$
	$\displaystyle \underbrace{=}_{{\rm Theorem}3.3-2}$	$\displaystyle -3\cos{x} + \frac{x^3}{3} + c\ensuremath{ \blacksquare}$

$\displaystyle \int \frac{1}{3+x^2}\: dx$	$\displaystyle =$	$\displaystyle \int \frac{1}{(\sqrt{3})^2+x^2}\: dx$
	$\displaystyle =$	$\displaystyle \frac{1}{\sqrt{3}}\tan^{-1}\big(\frac{x}{\sqrt{3}}\big) + c\ensuremath{ \blacksquare}$

Use $\int\frac{1}{a^2+x^2}dx = \frac{1}{a}\tan^{-1}{\frac{x}{a}} + c$ .

$\displaystyle \int \frac{1}{\sqrt{x^2 + 5}}\: dx$

$\displaystyle =$

$\displaystyle \log\vert x + \sqrt{x^2 + 5}\vert + c\ensuremath{ \blacksquare}$

Use $\int\frac{1}{\sqrt{x^2 + a}}dx = \log\vert x + \sqrt{x^2 + a}\vert + c$ .

Exercise 3..2 Integrate the following functions.
1. $\displaystyle{\frac{x^2 - 1}{x^2 + 1}}$ 2. $\displaystyle{\frac{x^2 + 2x - 1}{x^2 + 1}}$ 3. $\displaystyle{\frac{1}{\sqrt{4 - t^2}}}$ 4. $\displaystyle{\frac{1}{x^2 - 4}}$

SOLUTION

$\frac{\rm 2nd degree polynomial}{\rm 2nd degree polynomial} = {\rm costant} + \frac{\rm constant}{\rm 2nd degree polynomial}$ . 1.

$\displaystyle \int \frac{x^2 - 1}{x^2 + 1}dx$	$\displaystyle =$	$\displaystyle \int \frac{x^2 + 1 - 2}{x^2 + 1}dx = \int (1 - \frac{2}{x^2 + 1})dx$
	$\displaystyle =$	$\displaystyle x - 2\tan^{-1}{x} + c\ensuremath{ \blacksquare}$

2. $\int \frac{2x}{x^2 + 1}\:dx = \int \frac{(x^2+1)'}{x^2 + 1}\:dx = \log\vert x^2 + 1\vert + c$ . Now since $x^2 + 1 > 0$

, we write $\log(x^2 + 1)$ .

$\displaystyle \int \frac{x^2 + 2x - 1}{x^2 + 1}dx$	$\displaystyle =$	$\displaystyle \int \frac{x^2 + 1 + 2x - 2}{x^2 + 1}dx$
	$\displaystyle =$	$\displaystyle \int (1 + \frac{2x-2}{x^2 + 1})dx$
	$\displaystyle =$	$\displaystyle x + \int \frac{2x}{x^2+1}dx + \int\frac{-2}{x^2 + 1}dx$
	$\displaystyle =$	$\displaystyle x + \log(x^2 + 1) - 2\tan^{-1}{x} + c\ensuremath{ \blacksquare}$

3. $\int \frac{1}{\sqrt{a^2 -x^2}}dx = \sin^{-1}{\frac{x}{a}} + c$ .

$\displaystyle \int \frac{1}{\sqrt{4 - t^2}}\: dt$

$\displaystyle =$

$\displaystyle \int \frac{1}{\sqrt{2^2 - t^2}}\: dt = \sin^{-1}\big(\frac{t}{2}\big) + c\ensuremath{ \blacksquare}$

Write $\frac{1}{x^2 -4} = \frac{1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$ . Then find $A$ and $B$ . 4.

$\displaystyle \int \frac{1}{x^2 - 4}\: dx$	$\displaystyle =$	$\displaystyle \int \frac{1}{(x-2)(x+2)}\: dx = \int \frac{1}{2}\left(\frac{1}{x-2} - \frac{1}{x+2}\right)\: dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int \left(\frac{1}{x-2} - \frac{1}{x+2}\right)\:dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\left(\log\vert x-2\vert - \log\vert x+2\vert\right) + c = \frac{1}{2}\log\vert\frac{x-2}{x+2}\vert + c\ensuremath{ \blacksquare}$

Alternative solution By the rule of the integration, we have

$\displaystyle \int{\frac{1}{x^2 - 4}\:dx} = \int \frac{1}{x^2 - 2^2}\:dx = \log{\vert\frac{x-2}{x+2}\vert} + c$

Exercise A

1.

Work out the following integrals and check your answer by differentiation

(a) $\displaystyle{\int (2x-3) dx }$ (b) $\displaystyle{\int 5x^{4} dt}$ (c) $\displaystyle{\int \sqrt[3]{x^{2}} dx}$ (d) $\displaystyle{\int \frac{1}{\sqrt{x}} dx}$

(e) $\displaystyle{\int \frac{\sin{x}}{\cos{x}} dx}$ (f) $\displaystyle{\int \frac{1}{\cos^{2}x} dx}$ (g) $\displaystyle{\int \sin^{2}{x} dx}$ (h) $\displaystyle{\int \frac{1}{x^{2} - 4} dx}$

(i) $\displaystyle{\int \frac{1}{x^{2}+4}dx }$

Exercise B

1.

Work out the following integrals

(a) $\displaystyle{\int (3x^{-3} + 4x^{5}) dx }$ (b) $\displaystyle{\int (t^{3}+ \frac{1}{t^{2}+1}) dt}$ (c) $\displaystyle{\int -2\tan^{2}{x} dx}$

(d) $\displaystyle{\int \frac{x^{3} + 1}{\sqrt{x}} dx}$ (e) $\displaystyle{\int \frac{x^{2} + 5}{x^{2} + 4} dx}$ (f) $\displaystyle{\int \frac{1}{\sqrt{4 - t^{2}}} dt}$ (g) $\displaystyle{\int \cos^{2}{x} dx}$

(h) $\displaystyle{\int \frac{1}{\sqrt{t^{2} + 4}} dt}$ (i) $\displaystyle{\int \frac{1}{4 - x^{2}}dx }$