Antiderivatives

Primitive Functions Given a function $f(x)$ defined on some interval $I$. Then a function $F(x)$ safisfies

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

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

Understanding

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

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 $c$ 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 $G(x)$ and $F(x)$ be primitive functions of $f(x)$. Then $G'(x) = F'(x)$. This means that the slope of the tangent line to $G(x)$ and the slope of the tangent line to $F(x)$ is the same for all $x$. Thus the graph of $G(x)$ and $F(x)$ are parallet. Thus, the difference of $G(x)$ and $F(x)$ 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

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

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

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

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

Exercise 3..1   Integrate the following function

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

SOLUTION Note that

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

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

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

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

Check

Find $F(x)$ 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   $${1. \int x^{\alpha} dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c \hskip 0.3cm {\rm where} \alpha \neq -1}$$

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

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

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

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

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

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

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

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

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

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

$${12. \int \frac{dx}{\sqrt{x^{2} + a}} = \log{\vert x + \sqrt{x^{2} + a}\vert} + c}$$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. $${(\frac{x^{\alpha + 1}}{\alpha + 1} + c)^{\prime} = \frac{\alpha + 1}{\alpha + 1}x^{\alpha + 1 - 1} + c' = x^{\alpha}}$$. Thus, $${\int x^{\alpha} dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c}$$
10.

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

Rules of Integration

Theorem 3..3   Suppose that integral of $f(x)$ and $g(x)$ exist. Then

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

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

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

NOTE

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

$$= f(x) \pm g(x).$$

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

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

Thus, ,

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

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. $${3\sin{x} + x^2}$$ 2. $${\frac{1}{3 + x^2}}$$3. $${\frac{1}{\sqrt{x^2 + 5}}}$$

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

$$\int(3\sin{x} + x^2)dx \underbrace{=}_{{\rm Theorem}3.3-1} \int 3\sin{x} dx + \int x^2 dx$$

$$\underbrace{=}_{{\rm Theorem}3.3-2} 3\int \sin{x} dx + \int x^2 dx$$

$$\underbrace{=}_{{\rm Theorem}3.3-2} -3\cos{x} + \frac{x^3}{3} + c\ensuremath{ \blacksquare}$$

2.

$$\int \frac{1}{3+x^2}\: dx = \int \frac{1}{(\sqrt{3})^2+x^2}\: dx$$

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

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

3.

$$\int \frac{1}{\sqrt{x^2 + 5}}\: dx = \log\vert x + \sqrt{x^2 + 5}\vert + c\ensuremath{ \blacksquare}$$

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. $${\frac{x^2 - 1}{x^2 + 1}}$$ 2. $${\frac{x^2 + 2x - 1}{x^2 + 1}}$$ 3. $${\frac{1}{\sqrt{4 - t^2}}}$$ 4. $${\frac{1}{x^2 - 4}}$$

SOLUTION

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

$$\int \frac{x^2 - 1}{x^2 + 1}dx = \int \frac{x^2 + 1 - 2}{x^2 + 1}dx = \int (1 - \frac{2}{x^2 + 1})dx$$

$$= x - 2\tan^{-1}{x} + c\ensuremath{ \blacksquare}$$

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 $\log(x^2 + 1)$.

$$\int \frac{x^2 + 2x - 1}{x^2 + 1}dx = \int \frac{x^2 + 1 + 2x - 2}{x^2 + 1}dx$$

$$= \int (1 + \frac{2x-2}{x^2 + 1})dx$$

$$= x + \int \frac{2x}{x^2+1}dx + \int\frac{-2}{x^2 + 1}dx$$

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

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

$$\int \frac{1}{\sqrt{4 - t^2}}\: dt = \int \frac{1}{\sqrt{2^2 - t^2}}\: dt = \sin^{-1}\big(\frac{t}{2}\big) + c\ensuremath{ \blacksquare}$$

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.

$$\int \frac{1}{x^2 - 4}\: dx = \int \frac{1}{(x-2)(x+2)}\: dx = \int \frac{1}{2}\left(\frac{1}{x-2} - \frac{1}{x+2}\right)\: dx$$

$$= \frac{1}{2}\int \left(\frac{1}{x-2} - \frac{1}{x+2}\right)\:dx$$

$$= \frac{1}{2}\left(\log\vert x-2\vert - \log\vert x+2\vert\right) + c = \frac{1}{2}\log\vert\frac{x-2}{x+2}\vert + c\ensuremath{ \blacksquare}$$

Alternative solution By the rule of the integration, we have

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

Exercise A

1.

Work out the following integrals and check your answer by differentiation

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

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

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

Exercise B

1.

Work out the following integrals

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

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

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