7.2 Basic Integration Formulas

In Section 5.1, we learned a general rule for differentiation. To differentiate a function, we assume \(\Delta x\), calculate \(\Delta y\) and find the limit of \(\Delta y/\Delta x\) as \(\Delta x\to0\). However, integration has no general rule. To integrate we need to use our previous knowledge of the known results of differentiation. We have to answer this question: What function if differentiated will give the given function. This is a tentative process of finding our way back. To be able to integrate most functions in calculus, we do not need more than a few known integrals many of which we discuss in this section. The rest of the integrals can often be reduced to one of these known integrals by the following theorem and various methods that to be discussed in this chapter. At the end of this section, we present a small integral table that we often need.

Let’s start with a simple theorem.

Theorem 1. Let \(f(x)\) and \(g(x)\) be integrable functions, and \(k\) be a constant. Then \[\int kf(x)dx=k\int f(x)dx,\tag{a}\] i.e. the integral of the product of a constant and a function is equal to the product of the constant and the integral of that function. Also \[\int[f(x)\pm g(x)]dx=\int f(x)dx\pm\int g(x)dx,\tag{b}\] i.e., the integral of the sum or difference of two functions is the sum or difference of their integrals.

Show the proof

To prove the above relationships, we need to show the integral functions in the right hand side and the left hand side have the same derivatives (see Theorem 1 in section 7.1), and for this we need to use the properties of differentiation, and (d) in Section 7.1. For (a) \[\begin{aligned} \frac{d}{dx}\left(\int kf(x)dx\right) & =\frac{d}{dx}\left(k\int f(x)dx\right)\\ & =\left(\frac{dk}{dx}\right)\int f(x)dx+k\left(\frac{d}{dx}\int f(x)dx\right)\\ kf(x) & =0+kf(x).\end{aligned}\] For (b) \[\begin{aligned} \frac{d}{dx}\left(\int[f(x)\pm g(x)]dx\right) & =\frac{d}{dx}\left(\int f(x)dx\pm\int g(x)dx\right)\\ f(x)\pm g(x) & =\frac{d}{dx}\left(\int f(x)dx\right)\pm\frac{d}{dx}\left(\int g(x)dx\right)\\ f(x)\pm g(x) & =f(x)\pm g(x).\end{aligned}\]

Here it is assumed that the arbitrary constants are adjusted properly. Thus, the formula (b) asserts that the sum of any integral of \(f(x)\) and any integral of \(g(x)\) is an integral of \(f(x)+g(x)\).

We can combine the formulas (a) and (b) and write down for any finite number of functions

\[\begin{align}\bbox[#F2F2F2,5px,border:2px solid black]{ \int[a_{1}f_{1}(x)+\cdots+a_{n}f_{n}(x)]dx=a_{1}\int f_{1}(x)dx+\cdots+a_{n}\int f_{n}(x)dx.}\tag{c}\end{align}\]

Now let’s study the integrals of the some commonest functions. The results of Section 7.1 enable us to write down the following integrals

  1. \(\bbox[#F2F2F2,5px,border:2px solid black]{{\displaystyle \int\cos xdx=\sin x+C,}}\) 
  2. \(\bbox[#F2F2F2,5px,border:2px solid black]{{\displaystyle \int\sin xdx=-\cos x+C,}}\) 
  3. \(\bbox[#F2F2F2,5px,border:2px solid black]{{\displaystyle \int\frac{1}{1+x^{2}}dx=\arctan x+C.}}\)

For (2), we have applied example (3) of Section 7.1 and part (a) of the above theorem.

  • Some books drop the constant \(C\). In this case, the formulas must be understood as meaning that the function on the right-hand side is one integral function of the integrand (= function under the sign of integration).

 

Read the explanation for the integral of x⍺

Because for all values of \(r\in\mathbb{R}\) except \(r=0\), we have \[d(x^{r})=rx^{r-1}dx,\] or \[d\left(\frac{x^{r}}{r}\right)=x^{r-1}dx,\] it follows \[\int x^{r-1}dx=\frac{1}{r}x^{r}+C.\] Placing \(r-1=\alpha,\) we have \[\int x^{\alpha}dx=\frac{1}{\alpha+1}x^{\alpha+1}+C.\quad\quad(\alpha\in\mathbb{R},\ \alpha\neq-1)\]

Obviously when \(\alpha=-1\), this formula becomes meaningless (division by \(\alpha+1=0\) is not defined). We recall that \[\frac{d}{dx}\ln x=\frac{1}{x},\] but \(\ln x\) is defined only for \(x>0\). Thus \[\int\frac{dx}{x}=\ln x+C\qquad(\text{if }x>0)\] Next suppose \(x\) is negative. Then \(-x\) is positive, and hence \(\ln(-x)\) is defined. Also \[\frac{d}{dx}\ln(-x)=\frac{-1}{-x}=\frac{1}{x},\] so that, when \(x\) is negative, \[\int\frac{dx}{x}=\ln(-x)+C.\] We can unite the two formulas in one formula and write \[\int\frac{dx}{x}=\ln(\pm x)+C=\ln|x|+C,\] where we choose the ambiguous sign so that \(\pm x\) is positive: This formula holds for all real values of \(x\) other than \(x=0\).

Hence

 

  1. \(\bbox[#F2F2F2,5px,border:2px solid black]{{\displaystyle \int x^{\alpha}dx=\begin{cases} \dfrac{1}{\alpha+1}x^{\alpha+1}+C & (\alpha\neq-1)\\ \\ \ln|x|+C & (\alpha=-1) \end{cases}}}\)

This integral and formula (c) enable us to write down the integral of any polynomial: \[\int(a_{0}x^{n}+a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n})dx=\frac{a_{0}}{n+1}x^{n+1}+\frac{a_{1}}{n}x^{n}+\cdots+\frac{a_{n-1}}{2}x^{2}+a_{n}x+C.\] Here each of integrals has a constant of integration, say \(C_{1},\cdots,C_{n}\), and we combined them into one integration constant, i.e. \(C=C_{1}+\cdots+C_{n}\).

Recall that \[\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}}.\] [Some books denote the inverse of the sine function by \(\sin^{-1}x\) instead of \(\arcsin x\).] Thus, we have

  1. \(\bbox[#F2F2F2,5px,border:2px solid black]{{\displaystyle {\displaystyle \int\frac{dx}{\sqrt{1-x^{2}}}=\arcsin x+C.}}}\)

Because \[\frac{d}{dx}e^{x}=e^{x},\] we have

  1. \(\bbox[#F2F2F2,5px,border:2px solid black]{{\displaystyle {\displaystyle \int e^{x}dx=e^{x}+C.}}}\)
Example 1

Find \({\displaystyle \int\sqrt[3]{x}dx}\).

Solution 1

\[\begin{aligned} \int\sqrt[3]{x}dx=\int x^{1/3}dx & =\frac{1}{\frac{1}{3}+1}x^{1/3+1}+C\\ & =\frac{3}{4}x^{4/3}+C.\end{aligned}\]

Example 2

Find \({\displaystyle \int\frac{3d\theta}{\sqrt{\theta}}}\).

Solution 2

\[\begin{aligned} \int\frac{3}{\sqrt{\theta}}d\theta & =\int3\theta^{-\frac{1}{2}}d\theta\\ & =3\int\theta^{-\frac{1}{2}}d\theta\\ & =3\frac{1}{-\frac{1}{2}+1}\theta^{\frac{-1}{2}+1}+C\\ & =3(2)\theta^{\frac{1}{2}}+C\\ & =6\sqrt{\theta}+C.\end{aligned}\]

Example 3

Find \({\displaystyle \int\left(\frac{2a}{\sqrt{y}}-\frac{b}{y}+c\sqrt[3]{y^{2}}\right)}dy.\)

Solution 3

\[\begin{aligned} \int\left(\frac{2a}{\sqrt{y}}-\frac{b}{y}+c\sqrt[3]{y^{2}}\right)dy & =2a\int\frac{1}{\sqrt{y}}-b\int\frac{1}{y}dy+c\int\sqrt[3]{y^{2}}dy\\ & =2a\int y^{-1/2}dy-b\int y^{-1}dy+c\int y^{2/3}dy\\ & =2a\frac{1}{1-\frac{1}{2}}y^{-\frac{1}{2}+1}+b\ln|y|+c\frac{1}{1+\frac{2}{3}}y^{1+\frac{2}{3}}+C\\ & =4a\sqrt{y}+b\ln|y|+\frac{3c}{5}y^{\frac{5}{3}}+C.\end{aligned}\]

Example 4

Find \({\displaystyle \int\left(5\cos t+\frac{3t^{2}+\sqrt{t}}{t}\right)dt}\).

Solution 4

\[\begin{aligned} \int\left(5\cos t+\frac{3t^{2}+\sqrt{t}}{t}\right)dt & =5\int\cos t\ dt+3\int t\ dt+\int\frac{1}{\sqrt{t}}dt\\ & =5\sin t+\frac{3}{2}t^{2}+\frac{1}{-\frac{1}{2}+1}t^{-1/2+1}+C\\ & =5\sin t+\frac{3}{2}t^{2}+2\sqrt{t}+C.\end{aligned}\]

Example 5

Find \({\displaystyle \int\left[5(1+u^{2})^{-1}-e^{u}\right]du}\).

Solution 5

\[\begin{aligned} \int\left[5(1+u^{2})^{-1}-e^{u}\right]du & =5\int\frac{1}{1+u^{2}}du-\int e^{u}du\\ & =5\arctan u-e^{u}+C.\end{aligned}\]

Example 6

Find \({\displaystyle \int\left(\frac{x^{2}+3}{x}-\frac{2}{\sqrt{1-x^{2}}}\right)dx}\).

Solution 6

\[\begin{aligned} \int\left(\frac{x^{2}+3}{x}-\frac{2}{\sqrt{1-x^{2}}}\right)dx & =\int\left(x+\frac{3}{x}\right)dx-2\int\frac{1}{\sqrt{1-x^{2}}}dx\\ & =\frac{1}{2}x^{2}+3\ln|x|-2\arcsin x+C.\end{aligned}\]

Example 7

Find \({\displaystyle \int\frac{x^{2}}{1+x^{2}}dx}\).

Solution 7

\[\begin{aligned} \int\frac{x^{2}}{1+x^{2}}dx & =\int\frac{(x^{2}+1)-1}{1+x^{2}}dx\\ & =\int\left(1-\frac{1}{1+x^{2}}\right)dx\\ & =x-\arctan x+C.\end{aligned}\]

Example 8

Find \({\displaystyle \int\frac{x^{4}}{x^{2}+1}dx}\).

Solution 8

\[\begin{aligned} \int\frac{x^{4}}{x^{2}+1}dx & =\int\frac{x^{4}-1+1}{x^{2}+1}dx\\ & =\int\left(\frac{x^{4}-1}{x^{2}+1}+\frac{1}{x^{2}+1}\right)dx\\ & =\int\left(\frac{(x^{2}-1)(x^{2}+1)}{x^{2}+1}+\frac{1}{x^{2}+1}\right)dx\\ & =\int\left(x^{2}-1+\frac{1}{1+x^{2}}\right)dx\\ & =\frac{1}{3}x^{3}-x+\arctan x+C.\end{aligned}\]

Example 9

Find \({\displaystyle \int\sqrt{1-\cos2x}}\ dx\) for \(0\leq x\leq\pi\).

Solution 9

Because \((1-\cos2x)/2=\sin^{2}x\), we have \[\begin{aligned} \int\sqrt{1-\cos2x}\ dx & =\int\sqrt{2\sin^{2}x}\ dx\\ & =\sqrt{2}\int|\sin x|\ dx\\ & =\sqrt{2}\int\sin x\ dx\\ & =-\sqrt{2}\cos x+C.\end{aligned}\]

Note that \(\sin x\geq0\) when \(0\leq x\leq\pi\), so in this interval we have \(|\sin x|=\sin x\).

 

Table of Integration Formulas

  1. \({\displaystyle \int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx}\)
  2. \({\displaystyle \int kf(x)dx=k\int f(x)dx}\)
  3. \({\displaystyle \int x^{\alpha}dx=\frac{1}{\alpha+1}x^{\alpha+1}+C}\qquad\) (\(\alpha\neq-1)\)
  4. \({\displaystyle \int\frac{1}{x}dx=\ln|x|+C}\)
  5. \({\displaystyle \int\sin x}\ dx=-\cos x+C\)
  6. \({\displaystyle \int\cos x\ dx=\sin x+C}\)
  7. \({\displaystyle \int\sec^{2}x\ dx=\tan x+C}\)
  8. \({\displaystyle \int\sec x\tan x\ dx=\sec x+C}\)
  9. \({\displaystyle \int\csc^{2}x\ dx=-\cot x+C}\)
  10. \({\displaystyle \int\csc x\cot x\ dx=-\csc x+C}\)
  11. \({\displaystyle \int\frac{1}{x^{2}+1}dx=\arctan x+C}\)
  12. \({\displaystyle \int\frac{1}{\sqrt{1-x^{2}}}+C=\arcsin x+C}\)
  13. \({\displaystyle \int e^{x}dx=e^{x}+C}\)
  14. \({\displaystyle {\displaystyle \int\sinh x\ dx=\cosh x+C}}\)
  15. \({\displaystyle \int\cosh x\ dx=\sinh x+C}\)