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Next: 17.3 Integration by Parts Up: 17. Antidifferentiation Techniques Previous: 17.1 The Antidifferentiation Problem   Index

17.2 Basic Formulas

Every differentiation formula gives rise to an antidifferentiation formula. We review here a list of formulas that you should know. In each case you should verify the formula by differentiating the right side. You should know these formulas backward and forward.
$\displaystyle \int (f(x))^r f'(x)dx$ $\textstyle =$ $\displaystyle \frac{(f(x))^{r+1}}{r+1} \hspace{1em} (r \not = -1).\mbox{{}}$  
$\displaystyle \int \frac{f'(x)}{f(x)} dx$ $\textstyle =$ $\displaystyle \ln(\vert f(x)\vert).\mbox{{}}$  
$\displaystyle \int \cos(f(x))f'(x)dx$ $\textstyle =$ $\displaystyle \sin(f(x)). \mbox{{}}$  
$\displaystyle \int \sin(f(x))f'(x)dx$ $\textstyle =$ $\displaystyle -\cos(f(x)). \mbox{{}}$  
$\displaystyle \int e^{f(x)} f'(x) dx$ $\textstyle =$ $\displaystyle e^{f(x)}. \mbox{{}}$  
$\displaystyle \int \sec^2(f(x)) f'(x) dx$ $\textstyle =$ $\displaystyle \tan(f(x)). \mbox{{}}$  
$\displaystyle \int \csc^2(f(x)) f'(x) dx$ $\textstyle =$ $\displaystyle -\cot(f(x)). \mbox{{}}$  
$\displaystyle \int \sec(f(x))\tan(f(x)) f'(x) dx$ $\textstyle =$ $\displaystyle \sec(f(x)).\mbox{{}}$  
$\displaystyle \int \csc(f(x))\cot(f(x)) f'(x)dx$ $\textstyle =$ $\displaystyle -\csc(f(x)).\mbox{{}}$  
$\displaystyle \int \frac{f'(x)}{1 + f^2(x)} dx$ $\textstyle =$ $\displaystyle \arctan(f(x)). \mbox{{}}$  
$\displaystyle \int \frac{f'(x)}{\sqrt{1 - f^2(x)}}dx$ $\textstyle =$ $\displaystyle \arcsin(f(x)). \mbox{{}}
\index{antidifferentiation formulas}$  

17.7   Exercise. Verify that

\begin{displaymath}{d\over dx}\Big( \ln(\vert\sec(f(x)) + \tan(f(x))\vert) \Big)= \sec(f(x))f'(x)\end{displaymath}


\begin{displaymath}{d\over dx}\Big( \ln(\vert\csc(f(x)) + \cot(f(x))\vert)\Big) = -\csc(f(x))f'(x). \end{displaymath}

It follows from the previous exercise that

\begin{displaymath}\int \sec(f(x)) f'(x) dx = \ln(\vert \sec(f(x)) + \tan(f(x)) \vert) \end{displaymath}


\begin{displaymath}\int \csc(f(x)) f'(x) dx = -\ln(\vert\csc(f(x)) + \cot(f(x)) \vert). \end{displaymath}

You should add these two formulas to the list of antiderivatives to be memorized.

17.8   Theorem (Sum rule for antiderivatives) If $f$ and $g$ are functions that have antiderivatives on some interval $[a,b]$, and if $c \in \mbox{{\bf R}}$ then $f+g$, $f-g$ and $cf$ have antiderivatives on $[a,b]$ and

\begin{displaymath}\int (f\pm g) = \int f \pm \int g, \end{displaymath}

\begin{displaymath}\int cf = c \int f.
\end{displaymath} (17.9)

Proof:      The meaning of this statement is that if $F$ is an antiderivative for $f$ and $G$ is an antiderivative for $G$, then $F\pm G$ is an antiderivative for $f\pm g$, and $cF$ is an antiderivative for $cf$. The warning about ambiguous notation for indefinite integrals given in chapter 9 applies also to antiderivatives.

Let $F,G$ be antiderivatives for $f$ and $g$ respectively on $[a,b]$. Then $F$ and $G$ are continuous on $[a,b]$, and

\begin{displaymath}F' = f \mbox{ and }G' = g \end{displaymath}

on $(a,b)$. Hence $F\pm G$ are continuous on $[a,b]$, and

\begin{displaymath}(F\pm G)' = F' \pm G' = f \pm g \end{displaymath}

on $(a,b)$, and hence

\begin{displaymath}\int (f \pm g ) = F \pm G = \int f \pm \int g. \end{displaymath}

Also $cF$ is continuous on $[a,b]$, and

\begin{displaymath}(cF)' = cF' = cf\end{displaymath}

on $(a,b)$, so that

\begin{displaymath}\int cf = cF = c\int f.\mbox{ $\diamondsuit$}\end{displaymath}

17.10   Example. We will calculate $\displaystyle { \int x^2(x^3+1)^3 dx}$.

I will try to bring this integral into the form

\begin{displaymath}\int (f(x))^r f'(x) dx. \end{displaymath}

It appears reasonable to take $f(x) = (x^3+1)$, and then $f'(x) = 3x^2$. The $3x^2$ doesn't quite appear in the integral, but I can get it where I need it by multiplying by a constant, and using (17.9):

\begin{displaymath}\int x^2(x^3+1)^3 dx = \frac{1}{3} \int (3x^2) (x^3+1)^3 dx
= \frac{1}{3}\frac{(x^3+1)^4}{4} =\frac{(x^3+1)^4}{12}.\end{displaymath}

17.11   Example. We will calculate $\displaystyle { \int x(x^3+1)^3 dx}$.

This problem is more complicated than the last one. Here I still want to take $f(x) = (x^3+1)$, but I cannot get the ``$f'(x)$'' that I need. I will multiply out $(x^3+1)^3$

$\displaystyle \int x(x^3+1)^3 dx$ $\textstyle =$ $\displaystyle \int x( (x^3)^3 + 3(x^3)^2 + 3(x^3) + 1) dx \mbox{{}}$  
  $\textstyle =$ $\displaystyle \int (x^{10} + 3x^7 + 3x^4 + x)dx \mbox{{}}$  
  $\textstyle =$ $\displaystyle \frac{x^{11}}{11} + 3 \frac{x^8}{8} + 3 \frac{x^5}{5} + \frac{x^2}{2}.\mbox{{}}$  

17.12   Example. We will calculate $\displaystyle { \int t e^{t^2} dt}$.

\begin{displaymath}\int t e^{t^2} dt = \frac{1}{2} \int (2t) e^{t^2} dt. \end{displaymath}

Since $\displaystyle {\frac{d}{dt}(t^2) = 2t}$ we get

\begin{displaymath}\int te^{t^2} dt = \frac{1}{2} e^{t^2}.\end{displaymath}

17.13   Example. We will consider $\displaystyle { \int e^{t^2}dt}$.

Although this problem looks similar to the one we just did, it can be shown that no function built up from the functions we have studied by algebraic operations is an antiderivative for $\exp(t^2)$. So we will not find the desired antiderivative. (But by the fundamental theorem of the calculus we know that $e^{t^2}$ has an antiderivative.)

17.14   Example. We will calculate $\displaystyle {\int \tan }$.

\begin{displaymath}\int \tan = \int \frac{\sin}{\cos} = -\int\frac{\cos'}{\cos} = -\ln(\vert\cos\vert). \end{displaymath}

17.15   Example. We will calculate $\displaystyle {\int_0^a {1 \over a^2 + x^2}\; dx}$.

The integrand $\displaystyle {{1 \over a^2 + x^2}}$ looks enough like $\displaystyle {{1\over 1+x^2}}$ that I will try to get an $\arctan$ from this integral.

\begin{displaymath}\int {1 \over a^2 + x^2} = {1 \over a^2}
\int {1 \over 1 + \Big( {x\over a}\Big)^2}\; dx \end{displaymath}

Now $\displaystyle { {d\over dx}\left({x\over a}\right) = {1\over a}}$ , so

\begin{displaymath}\int {1 \over a^2 + x^2} = {1 \over a} \int {1 \over 1 + \Big...
...x\over a}\right)dx = {1\over a}\arctan\left( {x\over a}\right).\end{displaymath}

Thus we have found an antiderivative for $\displaystyle {{1 \over a^2 + x^2}}$. Hence

\begin{displaymath}\int_0^a {1 \over a^2 + x^2}\; dx =\left. {1\over a} \arctan(...
... a})
\right\vert _0^a
= {1\over a}\arctan(1) = {\pi \over 4a}.\end{displaymath}

17.16   Exercise. A Find the following antiderivatives:
$\displaystyle {\int e^x \sin(e^x) dx. }$
$\displaystyle {\int {e^x \over \sin(e^x)} dx. }$
$\displaystyle { \int (3w^4 + w)^2 (12w^3 + 1) dw. }$
$\displaystyle { \int \cos(4x) dx. }$
$\displaystyle { \int \frac{2x}{1+x^2} dx.}$
$\displaystyle { \int \cot(2x) dx.}$
$\displaystyle { \int \frac{2}{1+w^2} dw.}$
$\displaystyle { \int \frac{2w}{1+w^2} dw.}$
$\displaystyle { \int \sin^3 (x) dx.}$
$\displaystyle { \int \sin^4(x) dx.}$

next up previous index
Next: 17.3 Integration by Parts Up: 17. Antidifferentiation Techniques Previous: 17.1 The Antidifferentiation Problem   Index
Ray Mayer 2007-09-07