Let $f$ be any function and $F$ its antiderivative. The set of all antiderivatives of $f$ is called indefinite integral of the function $f$ denoted by
$$\int f(x) dx = F(x) + C.$$
Every two primitive functions differ by a constant $C$. This means, if we know the one antiderivative $F$ of the function $f$, then the another we can write in the form $F(x) + C$.
We do not have strictly rules for calculating the antiderivative (indefinite integral). The most antiderivatives we know is derived from the table of derivatives, which we read in the opposite direction.
Table of basic integrals
$$\int dx = x + C$$
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n\neq 1$$
$$\int \frac{1}{x} dx = \ln |x| + C$$
$$\int \frac{1}{x^n} dx = – \frac{1}{(n-1)x^{n-1}} + C, \quad n \neq 1$$
$$\int \frac{1}{\sqrt{x}} dx = 2 \sqrt{x} + C$$
$$\int e^x dx = e^x + C$$
$$\int a^x dx = \frac{a^x}{\ln a} + C, \quad a>0, a\neq 1$$
$$ \int \sin x dx = – \cos x + C$$
$$\int \cos x dx = \sin x + C$$
Table of integrals of rational functions
Example 1. Determine
$$\int (2 – 3x + x^2 )dx.$$
Solution.
By the additivity and linearity property of the integral, we have
$$\int (2 – 3x + x^2 )dx = 2 \int dx – 3 \int x dx + \int x^2 dx. $$
From the table of integral, we read
$$\int (2 – 3x + x^2 )dx = 2x – \frac{3}{2} x^2 + \frac{1}{3}x^3 + C.$$
Example 2. Determine
$$ \int \frac{x^3 -1}{2x^2} dx.$$
Solution.
$$\int \frac{x^3 -1}{2x^2} dx = \int \left(\frac{x^3}{2x^2} – \frac{1}{2x^2} \right) dx $$
$$= \int \left(\frac{x}{2} – \frac{1}{2x^2} \right) dx$$
$$ = \frac{1}{2} \left( \int x – \frac{1}{x^2} \right) dx.$$
By the additivity property of the integral, we have
$$\frac{1}{2} \left( \int x – \frac{1}{x^2} \right) dx = \frac{1}{2} \left (\int x dx – \int \frac{1}{x^2} dx \right). $$
From the table of indefinite integrals, we read:
$$\frac{1}{2} \left (\int x dx – \int \frac{1}{x^2} dx \right) = \frac{1}{2} \left ( \frac{1}{2}x^2 + \frac{1}{x} \right) + C $$
$$ = \frac{x^2}{4} + \frac{1}{2x} + C,$$
that is
$$\int \frac{x^3 -1}{2x^2} dx = \frac{x^2}{4} + \frac{1}{2x} + C.$$
Example 3. Determine
$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx .$$
Solution.
Since $\sin 2x = 2 \sin x \cos x$, we can write
$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx = \int \frac{2 \sin x \cos x + 3 \sin^2 x}{\sin x} dx $$
$$ = \int \left( \frac{2 \sin x \cos x}{\sin x} + \frac{3 \sin^2 x}{\sin x} \right) dx$$
$$= \int \left( 2 \cos x + 3 \sin x \right) dx.$$
By the linearity and additivity property of the integral, we have
$$\int \left( 2 \cos x + 3 \sin x \right) = 2 \int \cos x dx + 3 \int \sin x dx.$$
From the table of indefinite integrals we read:
$$ 2 \int \cos x dx + 3 \int \sin x dx = 2 \sin x – 3 \cos x + C,$$
that is
$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx = 2 \sin x – 3 \cos x + C.$$
Integration by substitution
We will use the reverse chain rule of differentiation of composite functions and thus obtain the method which is called the method of substitution.
The chain rule:
$$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x).$$
The reverse of the chain rule:
$$\int f’ (g(x)) g'(x) dx = f(g(x)) + C.$$
The integration rule: if $g: I \to \mathbb{R}, I \subseteq \mathbb{R}$ is a differentiable and continuous function on $I$, then
$$\int f(g(x)) g'(x) dx = \int f(u) du,$$
where $g(x) = u$ and $g'(x) dx = du$.
Firstly, we need to choose a substitution to make. A substitution is not determined in advance, just by using the exercises we can discover the simplest way. Then we have to express the variable $x$ by $u$ and connect $du$ and $dx$. We change the integrand and $dx$ with expressions by the variable $u$ and calculate the integral. The result we write as the function of the variable $x$, that is, we return a substitution.
Example 4. Determine
$$ \int \sqrt{3 + x} dx.$$
Solution.
We choose a substitution $u= \sqrt{3+x}, (u \ge 0).$ It follows
$$x= u^2 – 3$$
and
$$dx = 2u du.$$
Now, by changing $x$ with $u$, we have:
$$\int \sqrt{3 + x} dx = \int u 2u du = \int 2u^2 du.$$
From the linearity property of integrals we can write
$$ \int 2u^2 du = 2 \int u^2 du.$$
From the table of integrals we read:
$$2 \int u^2 du = \frac{2}{3} u^3 + C.$$
Finally, the result we need to write as the function of the variable $x$:
$$\frac{2}{3} u^3 + C = \frac{2}{3} \sqrt{(3+x)^3} + C,$$
that is
$$\int \sqrt{3 +x } dx = \frac{2}{3} \sqrt{(3+x)^3} + C.$$
Example 5. Determine
$$\int \frac{2x + 1}{x^2 + x -3} dx.$$
Solution.
We choose a substitution $u = x^2 + x – 3$. Therefore, we have
$$\frac{du}{dx} = 2x +1 \Rightarrow dx = \frac{du}{2x+1}.$$
By replacing $x^2 + x – 3$ with $u$ and $dx$ with $\frac{du}{2x+1}$ we obtain:
$$\int \frac{2x + 1}{x^2 + x -3} dx = \int \frac{2x+1}{u} \cdot \frac{du}{2x + 1} = \int \frac{du}{u}.$$
From the table of integrals we read:
$$\int \frac{du}{u} = \ln |u|+ C.$$
By returning a substitution, we obtain the finally result
$$ \int \frac{2x + 1}{x^2 + x -3} dx = \ln |x^2 + x – 3| + C.$$
Example 6. Determine
$$\int \frac{\ln x}{x} dx.$$
Solution.
We choose a substitution $u = \ln x$. It follows
$$\frac{du}{dx} = \frac{1}{x} \Rightarrow dx = x du.$$
By replacing $\ln x$ with $u$ and $dx$ with $x du$, we have
$$\int \frac{\ln x}{x} dx = \int \frac{u}{x} x du = \int u du.$$
From the table of integrals we read:
$$\int u du = \frac{1}{2} u^2 + C,$$
that is
$$\int \frac{\ln}{x} dx = \frac{1}{2} \ln^2 x + C.$$
Example 7. Determine
$$\int \cot x dx.$$
Solution.
Since $\cot x = \frac{\cos x}{\sin x}$, we can write
$$\int \cot x dx = \int \frac{\cos x}{\sin x} dx.$$
We choose a substitution $u = \sin x$. It follows
$$\frac{du}{dx} = \cos x \Rightarrow dx = \frac{du}{\cos x}.$$
Therefore, we have
$$\int \frac{\cos x}{\sin x} dx = \int \frac{\cos x}{u} \cdot \frac{du}{\cos x} = \int \frac{du}{u}.$$
From the table of integrals we read
$$\int \frac{du}{u} = \ln|u| + C.$$
By returning a substitution, we obtain
$$\int \cot x dx = \ln |\sin x| + C.$$
The same principle we use for calculating the definite integral. However, the result we do not need write as the function of the variable $x$, because the result is a number. Instead, together with the integrand we are changing the lower and upper limit of integration.
Example 8. Determine
$$\int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^3 x dx.$$
Solution.
We choose $u = \cos x$. It follows
$$\frac{du}{dx} = – \sin x \Longrightarrow dx = – \frac{du}{\sin x}.$$
The lower and upper limit are transformed in the following way:
$$x = 0 \Longrightarrow u = \cos 0 = 1,$$
$$x = \frac{\pi}{2} \Longrightarrow u = \cos \frac{\pi}{2} = 0.$$
Now we have:
$$\int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^3 x dx = \int_{0}^{1}u^4 \cdot \sin^3 x \cdot \left( – \frac{du}{\sin x} \right) $$
$$= \int_{0}^{1} u^4 \cdot (-\sin^2) du.$$
From the relation $\sin^2 x + \cos ^2 x = 1$, we can write $\sin^2 x = 1 – \cos^2 x$, that is
$$ \int_{0}^{1} u^4 \cdot (-\sin^2) du = \int_{0}^{1} u^4 \cdot (-1)(1 – \cos^2 x) du $$
$$= \int_{0}^{1} u^4 \cdot (u^2 -1) du$$
$$ = \int_{0}^{1} (u^6 – u^4) du$$
$$= \int_{0}^{1} u^6 du – \int_{0}^{1} u^4 du.$$
By using the first fundamental theorem of calculus, finally we have:
$$\int_{0}^{1} u^6 du – \int_{0}^{1} u^4 du = \frac{u^7}{7} \bigg|_{0}^{1} – \frac{u^5}{5}\bigg|_{0}^{1} = – \frac{2}{35}$$
Integration by parts
We cannot calculate all integrals by using the method of substitution. Therefore, for calculate the simple integral as $\int x \ln x$ we need to use different methods.
Integration by parts is the one useful method for calculating integrals. Let $u$ and $v$ be functions of the variable $x$. The formula for derivative of the product
$$[u(x)v(x)]’ = u'(x)v(x) + u(x)v'(x)$$
can be written in the equivalent form:
$$u(x)v(x) = \int [u'(x)v(x) + u(x)v'(x)] dx = \int u'(x) v(x) dx + \int u(x)v'(x) dx.$$
Thus we obtain:
$$\int u(x)v'(x) dx = u(x)v(x) – \int u'(x)v(x) dx, $$
or more simply
$$\int u v’ dx = uv – \int u’ v dx.$$
The formula above is called the formula for integration by parts.
The method of the integration by parts
- The integrand write as the product of functions $u$ and $v’$.
- Calculate the extra integral $v= \int v’ dx$. The function $v’$ must be chosen in the way that its integral is easy to determine.
- Calculate the derivative $u’$.
- Write the formula for integration by parts:
$$\int uv’ dx = uv – \int u’v dx.$$
- Calculate the integral $\int u’v dx$.
Example 9. Determine
$$\int x \sin x dx.$$
Solution.
We choose $v’ = \sin x dx$ and $u = x$. Now we calculate the integral $v= \int v’ dx$, that is
$$v’ = \sin x \Rightarrow v= \int \sin x dx = – \cos x + C.$$
The derivative of $u$ is equal to
$$u = x \Rightarrow u’ = 1.$$
By using the formula for integration by parts we have
$$\int x \sin x dx =x \cdot( – \cos x) + \int \cos x.$$
From the table of integrals we read: $\int \cos x = \sin x + C$. Finally, we have
$$\int x \sin x dx = – x \cdot \cos x + \sin x + C.$$
In some cases, integration by parts must be repeated.
Example 10. Determine
$$\int e^x \cos x dx.$$
Solution.
We choose $u = e^x$ and $v’ = \cos x dx$. It follows
$$v = \int \cos x dx \Longrightarrow v = \sin x ,$$
$$du = e^x dx.$$
By using the formula for integration by parts, we have:
$$\int e^x \cos x dx =e^x \cdot \sin x – \int e^x \sin x dx.$$
Now we choose $t = e^x$ and $w’ = \sin x dx$. It follows
$$w = \int \sin x dx \Longrightarrow w = – \cos x ,$$
$$dt = e^x dx,$$
that is
$$\int e^x \cos x dx =e^x \cdot \sin x – \int e^x \sin x dx$$
$$= e^x \cdot \sin x – \left[e^x \cdot ( – \cos x) – \int e^x \cos x dx \right]$$
$$ = e^x \cdot \sin x + e^x \cdot \cos x – \int e^x \cos x dx.$$
Finally, we have
$$2 \int e^x \cos x dx = e^x ( \sin x + \cos x) + C,$$
that is
$$\int e^x \cos x dx = \frac{1}{2} e^x ( \sin x + \cos x) + C.$$
For the definite integral is valid the formula for integration by parts:
$$\int_{a}^b u(x) v'(x) dx = u(x) v(x) \bigg|_{a}^{b} – \int_{a}^{b} u'(x) v(x) dx.$$
Example 11. Determine
$$\int_{2}^{3} x e ^{x} dx.$$
Solution.
We choose $u= x$ and $v’= e^x dx$. It follows
$$v = \int e^x dx \Longrightarrow v = e^x,$$
$$du =dx.$$
By using the formula for integration by parts for definite integral and first fundamental theorem of calculus, we have:
$$\int_{2}^{3} x e ^{x} dx = x \cdot e^x \bigg|_{2}^3 – \int_{2}^{3} e^x dx $$
$$= x \cdot e^x \bigg|_{2}^3 – e^x \bigg|_2^3 $$
$$=(3 \cdot e^3 – 2 \cdot e^2) – (e^3 – e^2) $$
$$=2 e^3 – e^2 $$
$$=e^2(2e -1).$$
