This online calculator will calculate the integral of any function.
When the integrand matches a known form, it applies fixed rules to solve the integral
(e. g. partial fraction decomposition for rational functions, trigonometric substitution for integrands involving the square roots of a
quadratic polynomial or integration by parts for products of certain functions).
Otherwise, it tries different substitutions and transformations until either the integral is solved, time runs out or there is nothing left to try.
The calculator lacks the mathematical intuition that is very useful for finding an antiderivative,
but on the other hand it can try a large number of possibilities within a short amount of time.
It can show the steps and graphing for both input and result function.
It can handle polynomial, rational, irrational, exponential, logarithmic, trigonometric, inverse trigonometric, hyperbolic and inverse hyperbolic functions.
You can pick a random function using the 'Random function' button.
- trig
- Shows the trigonometry functions.
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- del
- Deletes the last element before the cursor.
- clear
- Removes all the text in the textfield.
- log
- logarithm (base e)
- pi
- Mathematical constant pi
- e
- Mathematical constant e
- ^
- Power
- sqrt
- Square root
Consider a function \(f\left( x \right) = 6x^{4} - 7x^{2} + 6\). The
process of getting its slope, the derivative, leads to another function
\(g\left( x \right) = 24x^{3} - 14x\). We denote \(g(x)\) by
\(\frac{d}{\text{dx}}\left( f\left( x \right) \right)\) and call it the
derivative of the function \(f\left( x \right).\) However, there are
situations where we may need to get back the original function. You
notice that from the derivative function,
\(g\left( x \right) = 24x^{3} - 14x\), we get the original function by
applying a transformation from \(ax^{n}\) to
\(\frac{ax^{n + 1}}{n + 1}\) to get
\(f\left( x \right) = \frac{24x^{3 + 1}}{3 + 1} - \frac{14x^{1 + 1}}{1 + 1} = 6x^{4} - 7x^{2}.\)
For the sake of the constant value, \(6\), we add a general constant
symbol, \(C\) to represent any constant. Thus, we get
\(f\left( x \right) = 6x^{4} - 7x^{2} + C.\) The process of getting the
original function from its derivative is called
\(\mathbf{\text{integration}}\) and the function obtained after
integration is called an integral.
Therefore, if the \(y = f(x)\) is a differentiable function, such that
its derivative is \(g(x)\), then the integral of \(g(x)\) with respect
to the variable, \(x\), is \(\text{y.}\) And we write
\[\frac{df(x)}{\text{dx}} = g(x)\]
implying
\[df(x) = g\left( x \right)\text{dx}\]
Hence, applying the integral sign on both sides, we get
\[y = f\left( x \right) = \int_{}^{}{df(x)} = \int_{}^{}{g\left( x \right)\text{dx}} + C\]
where \(C\) is a constant of integration.
In the integral representation,
\(\int_{}^{}{g\left( x \right)\text{dx}}\) the function \(g(x)\) to be
integrated is called the integrand. The integral of a function
is also called the anti-derivative. Thus,
\(\int_{}^{}{g\left( x \right)\text{dx}} + C\) is the antiderivative of
\(g\left( x \right)\) or an indefinite integral. If a function has an
antiderivative, we say it is integrable.
Example
Find the anti-derivative of
\(\mathbf{g}\left( \mathbf{x} \right)\mathbf{= 9}\mathbf{x}^{\mathbf{2}}\mathbf{+ 5}\mathbf{x + 1}\)
Solution -
The antiderivative of \(g(x)\) is
\[y = \int_{}^{}{g(x)}\ dx = \int_{}^{}{\left( 9x^{2} + 5x + 1 \right)\text{dx}} = \frac{9x^{2 + 1}}{2 + 1} + \frac{5x^{1 + 1}}{1 + 1} + \frac{1x^{0 + 1}}{0 + 1} + C\]
\[= \frac{9x^{3}}{3} + \frac{5x^{2}}{2} + \frac{x^{1}}{1} + C\]
\[= 3x^{3} + \frac{5x^{2}}{2} + x + C\]
If the functions \(g(x)\) and \(f(x)\) are integrable and \(k\) is a constant, then
(i).
\(\int_{}^{}{\lbrack g\left( x \right) \pm f(x)\rbrack}dx = \int g\left( x \right)dx \pm \int f\left( x \right)\text{dx}\)
(ii). \(\int_{}^{}{k\ g(x)}dx = k\int g\left( x \right)\text{dx}\)
(iii). \(\int_{}^{}kdx = k\int dx = kx + C\)
(iv). \(\int_{}^{}x^{n}\ dx = \frac{x^{n + 1}}{n + 1} + C\)
These properties are very basic, hence, in most cases, they are
overlooked. For instance, the integral of \(9x^{2} + 5x + 1\) above is
given by \(3x^{3} + \frac{5x^{2}}{2} + x + C.\) If we follow all the
steps based on the properties of integral, we will have the following
\[\int_{}^{}{\left( 9x^{2} + 5x + 1 \right)\text{dx}} = \int_{}^{}{9x^{2}}\ dx + \int_{}^{}{5x} + \int_{}^{}{1\ dx}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 9\int_{}^{}x^{2}\ dx + 5\int_{}^{}x + 1\int_{}^{}\text{dx}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 9 \cdot \frac{x^{2 + 1}}{2 + 1} + 5 \cdot \frac{x^{1 + 1}}{1 + 1} + 1 \cdot \frac{x^{0 + 1}}{0 + 1} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 9 \cdot \frac{x^{3}}{3} + 5 \cdot \frac{x^{2}}{2} + 1 \cdot \frac{x^{1}}{1} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3x^{3} + \frac{5x^{2}}{2} + x + C\]
There are definite integrals of involving integration of infinite
intervals or integrating on intervals having vertical asymptotes
a). Integration over infinite interval
If \(f\left( x \right)\) is an integrable function then
$$
\int_{-\infty}^{\infty} f(x) d x=\int_{-\infty}^{0} f(x) d x+\int_{0}^{\infty} f(x) d x=\lim _{t \rightarrow-\infty} \int_{t}^{0} f(x) d x+\lim _{s \rightarrow \infty} \int_{0}^{s} f(x) d x
$$
Example
Evaluate the integral
\[\int_{1}^{\infty\ }\frac{1}{x^{4}}\text{dx}\]
Solution
$$
\begin{aligned}
\int_{1}^{\infty} \frac{1}{x^{4}} d x &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{x^{4}} d x \\
&=\lim _{t \rightarrow \infty} \int_{1}^{t} x^{-4} d x \\
&=\lim _{t \rightarrow \infty}\left[\frac{x^{-3}}{-3}\right]_{1}^{t} \\
&=-\frac{1}{3} \lim _{t \rightarrow \infty}\left(t^{-3}-1^{-3}\right) \\
&=-\frac{1}{3} \lim _{t \rightarrow \infty}\left(\frac{1}{t^{3}}-1\right)=\frac{1}{3}
\end{aligned}
$$
b). Integration near a vertical asymptote
Example
Evaluate
\[\int_{0}^{\pi/4}{\tan\left( 2x + \frac{\pi}{2} \right)\operatorname{\ sec}\left( 2x + \frac{\pi}{2} \right)\text{dx}}\ \]
Let \(u = 2x + \frac{\pi}{2}\) then \(du = 2\ dx\) or
\(\frac{\text{du}}{2} = dx\). When \(x = 0,\ u = \frac{\pi}{2}\) and
when
\(x = \frac{\pi}{4},\ u = 2\left( \frac{\pi}{4} \right) + \frac{\pi}{2} = \pi.\)
Upon substitution, we have
\[\int_{0}^{\pi/4}{\tan\left( 2x + \frac{\pi}{2} \right)\operatorname{\ sec}\left( 2x + \frac{\pi}{2} \right)\text{dx}} = \lim_{a \rightarrow \frac{\pi}{2}^{+}\ }{\int_{a}^{\pi}{\tan u\operatorname{\ sec}u\frac{\text{du}}{2}}}\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \lim_{a \rightarrow \frac{\pi}{2}^{+}\ }\frac{1}{2}\int_{a}^{\pi}{\tan u\operatorname{\ sec}u\text{du}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\lim_{a \rightarrow \frac{\pi}{2}^{+}\ }\int_{a}^{\pi}{\tan u\operatorname{\ sec}u\text{du}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\lim_{a \rightarrow \frac{\pi}{2}^{+}\ }\left\lbrack \sec u \right\rbrack_{a}^{\pi}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\lim_{a \rightarrow \frac{\pi}{2}^{+}\ }\left( \sec\pi - \sec a \right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\left( \sec\pi - \lim_{a \rightarrow \frac{\pi}{2}^{+}\ }\sec a \right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\left( - 1 - \lim_{a \rightarrow \frac{\pi}{2}^{+}\ }\sec a \right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\left( - 1 - ( - \infty) \right)\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = does\ not\ converge\]
Therefore, the integral does not converge
Given the integral
\[\int_{}^{}{f\left( x \right)\text{dx}}\]
Where we can find \(u\left( x \right),\) then we can do substitution for
the expression \(u(x)\) and for \(dx = du(x)\) so that we get a simple
function that we can easily integrate.
Example
Solve the integral
\[\int_{}^{}{2x\left( 3x^{2} + 5 \right)^{3}\text{dx}}\]
Solution
Let \(u = 3x^{2} + 5\) then \(du = 6x\ dx\) or
\(\frac{\text{du}}{3} = 2x\ dx\)
Upon substitution, we get
\[\int_{}^{}{2x\left( 3x^{2} + 5 \right)^{3}\text{dx}}\mathbf{=}\int_{}^{}{u^{3}\ \frac{\text{du}}{3}}\]
\[\mathbf{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =}\frac{1}{3}\int_{}^{}u^{3}\text{ du}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \cdot \frac{\text{ u}^{3}}{3} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{6}\left( u^{3} \right) + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{6}\left( 3x^{2} + 5 \right)^{3} + C\]
Example
Solve the integral
\[\int_{0}^{2}{\ 4x\sqrt{{2x}^{2} + 1}\text{dx}}\text{}\]
Solution
Let \(u = 2x^{2} + 1\) then \(du = 4x\ dx\). When \(x = 2,\)
\(u = 2\left( 2 \right)^{2} + 1 = 9\) and when \(x = 0\), \(u = 1.\)
Upon substitution we get
\[\int_{\mathbf{0}}^{\mathbf{2}}{\mathbf{\ }4x\sqrt{{2x}^{2} + 1}\text{dx}}\mathbf{=}\int_{\mathbf{1}}^{\mathbf{9}}{\mathbf{\ }\sqrt{u}\text{du}}\]
\[\mathbf{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =}\int_{\mathbf{1}}^{\mathbf{9}}{\mathbf{\ }u^{\frac{1}{2}}\text{du}}\]
\[\mathbf{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =}\left\lbrack \frac{2}{3}u^{\frac{3}{2}} \right\rbrack_{\mathbf{1}}^{\mathbf{9}}\]
\[\mathbf{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =}\frac{2}{3}\left\lbrack \left( \sqrt{u} \right)^{3} \right\rbrack_{1}^{9}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{2}{3}\left\lbrack \left( \sqrt{9} \right)^{3} - \left( \sqrt{1} \right)^{3} \right\rbrack\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{2}{3}\left\lbrack 27 - 1 \right\rbrack\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{2}{3}\left( 26 \right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{52}{3}\]
The integral
\[\int_{}^{}{\frac{u'(x)}{u(x)}\text{dx}} = \ln{u(x)} + C\]
Example
\[\int_{}^{}{\frac{18x + 24}{\left( 3x + 4 \right)^{2}}\text{dx}}\]
Solution
Let \(u = \left( 3x + 4 \right)^{2}\) then
\(du = 2\left( 3x + 4 \right) \cdot 3\ dx = 6\left( 3x + 4 \right)\text{dx}\).
Then
\[\int_{}^{}{\frac{18x + 24}{\left( 3x + 4 \right)^{2}}\text{dx}} = \int_{}^{}{\frac{6\left( 3x + 4 \right)}{\left( 3x + 4 \right)^{2}}\text{dx}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\frac{1}{u}\text{du}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ln\left| u \right| + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ln\left| \left( 3x + 4 \right)^{2} \right| + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ln\left( 3x + 4 \right)^{2} + C\]
The rules for integration of simple trigonometric functions is given below
\[\int_{}^{}{\sin x\text{dx}} = \ - \cos x + C\]
\[\int_{}^{}{\cos x\text{dx}} = \sin x + C\]
\[\int_{}^{}{\sec^2{x}\text{dx}} = \ \tan x + C\]
\[\int_{}^{}{\csc^2{x}\text{dx}} = \ - \cot x + C\]
\[\int_{}^{}{\csc x\cot x\text{dx}} = \ - \csc x + C\]
\[\int_{}^{}{\sec x\tan x\text{dx}} = \ \sec x + C\]
Integral of \(\sec x,\) \(\csc x\), \(\tan x\) and \(\cot x\) requires
more manipulations.
For instance
\[\int_{}^{}{\tan x\text{dx}} = \int_{}^{}{\frac{\sin x}{\cos x}\text{dx}}\]
Let \(u = \cos x\) then \(du = - \sin x\text{dx}\) or
\(- du = \sin x\text{dx}\)
\[\int_{}^{}{\frac{\sin x}{\cos x}\text{dx}} = \int_{}^{}\frac{- du}{u}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \int_{}^{}\frac{\text{du}}{u} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ - \ln\left| u \right| + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \ln\left| \cos\text{x } \right| + C\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ln\left| \frac{1}{\cos x} \right| + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ln{\left| \sec x \right| + C}\]
Given that a function \(f(x)\) is differentiable and
\(g\left( x \right) = h\left( f\left( x \right) \right)\) where \(h(x)\)
is a trogonpmetric function, we can use the substitution \(u = f(x)\)
and \(g\left( u \right) = h(u)\) then do the integration.
Example
\[\int_{0}^{\pi/16}{\sec^2\left( 4x \right)\text{dx}}\]
Solution
Let \(u = 4x\) then \(du = 4 dx\ \) and
\(dx = \frac{1}{4}\text{du.}\) When \(x = 0,\ u = 0\) and when
\(x = \frac{\pi}{16},\ u = \frac{\pi}{4}\). Upon substitution, we have
\[\int_{0}^{\frac{\pi}{16}}{\sec^2\left( 4x \right)\text{dx}} = \frac{1}{4}\int_{0}^{\frac{\pi}{4}}{\sec^2\left( u \right)\text{du}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{4}\left\lbrack \tan u \right\rbrack_{0}^{\frac{\pi}{4}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{4}\left( \tan{\left( \frac{\pi}{4}\ \right) - \tan 0} \right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{4}\left( 1 - 0 \right)\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{4}.\]
When we have the integrand as a product of two trigonometric, we use
substitution method. The function \(u(x)\) is selected suitably so that
remaining part if a faction of \(u\) that is polynomial.
Example
Integrate \(\int_{}^{}{\cos^2{x}\sin^3{x}\text{dx}}\)
Solution
Here, we take \(u(x)\) to be the function with the lowest power.
\[\int_{}^{}{\cos^2{x}\sin^3{x}\text{dx}} = \int_{}^{}{\cos^2{x}\sin^2{x} \cdot \sin x\text{dx}}\]
Let \(u = cos x\), then \(du = \sin x\text{dx}\)
We now write \(\sin x\) in terms of cos \(x\) by applying the
trigonometric identity \(\sin^2{x} + \cos^2{x} = 1\ \) so
that \(\sin^2{x} = 1 - \cos^2{x} = 1 - u^{2}\). Then
\[\int_{}^{}{\cos^2{x}\sin^2{x} \cdot \sin x\text{dx}} = \int_{}^{}{u^{2}\left( 1 - u^{2} \right)\text{du}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\left( u^{2} - u^{4} \right)\text{du}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{u^{3}}{3} - \frac{u^{5}}{5} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\cos^3{x}}{3} - \frac{\cos^5{x}}{5} + C\]
Example
Integrate \(\int_{}^{}{\sin^9{x}\text{dx}}\)
Solution
\[\int_{}^{}{\sin^9{x}\text{ dx}} = \int_{}^{}{\sin^8{x}\sin x\text{dx}}\]
Let \(u = \cos x\) then \(du = \sin x\text{dx}\). Using the
trigonometric identity \(\sin^2{x} + \cos^2{x} = 1\ \) so
that \(\sin^2{x} = 1 - \cos^2{x} = 1 - u^{2}\). Then
\[\int_{}^{}{\sin^8{x}\sin x\text{dx}} = \int_{}^{}\left( 1 - u^{2} \right)^{4}\text{du}\]
Using binomial expansion, we have
\[\int_{}^{}\left( 1 - u^{2} \right)^{4}\ du = \int_{}^{}\left( 1 - 4\left( u^{2} \right) + 6\left( u^{2} \right)^{2} - 4\left( u^{2} \right)^{3} + \left( u^{2} \right)^{4} \right)\text{du}\]
\[\ \ = \int_{}^{}\left( 1 - 4u^{2} + 6u^{4} - 4u^{6} + u^{8} \right)\text{du}\]
\[= u - \frac{4}{3}u^{3} + \frac{6}{5}u^{5} - \frac{4}{7}u^{7} + \frac{u^{9}}{9} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = u - \frac{4}{3}\sin^3{x} + \frac{6}{5}\sin^5{x} - \frac{4}{7}\sin^7{x} + \frac{1}{9}\sin^9{x} + C\]
When we have the integrand as a quotient of two trigonometric, we use
substitution method. The function \(u(x)\) is selected suitably so that
remaining part if a faction of \(u\) that is polynomial.
Example
Evaluate
\[\int_{}^{}\frac{\sec^2{x}}{\tan^3{x}}\text{dx}\]
Solution
Let \(u = \tan x\), then \(du = \sec^2{x}\text{dx}\) so that
\[\int_{}^{}\frac{\sec^2{x}}{\tan^3{x}}\ dx = \int_{}^{}\frac{\text{du}}{u^{3}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{u^{- 3}\text{\ du}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{u^{- 3 + 1}}{- 3 + 1} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{1}{2}u^{- 2}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{1}{2u^{2}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{1}{2\tan^2{x}} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{1}{2}\cot^2{x} + C\]
When an integral involves a denominator or even the numerator is one of
the following root, then a corresponding substitution is done to solve
it.
a). \(a^{2} - b^{2}x^{2}\) take \(x = \frac{a}{b}\sin t\) then use the
identity \(1 - \sin^2{t} = \cos^2{t}\)
b). \(a^{2}x^{2} - b^{2}\) take \(x = \frac{a}{b}\sec t\) then use the
identity \(\sec^2{t} - 1 = \tan^2{t}\)
c). \(a^{2} + b^{2}x^{2}\) take \(x = \frac{a}{b}\tan t\) then use the
identity \(1 + \tan^2{t} = \sec^2{t}\)
Example
Integrate
\[\int_{}^{}\frac{\text{dx}}{\sqrt{4 - 9x^{2}}}\]
Solution
This is of the form \(a),\) hence, let \(x = \frac{2}{3}\sin t\), then
\(x^{2} = \frac{4}{9}\sin^2{t}\) and
\(dx = \frac{2}{3}\cos t\text{dt}\).
From \(\frac{2}{3}\sin t = x\), we have
\(t = \sin^{-1}\left( \frac{3x}{2} \right)\ \)
Upon substitution, we have
\[\int_{}^{}\frac{\text{dx}}{\sqrt{4 - 9x^{2}}} = \int_{}^{}{\frac{\frac{2}{3}\cos t\text{dt}}{\sqrt{4 - 9 \cdot \left( \frac{4}{9}\sin^2{t} \right)}}\ } = \int_{}^{}{\frac{\frac{2}{3}\cos t\text{dt}}{\sqrt{4 - 4\sin^2{t}}}\ }\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\frac{\frac{2}{3}\cos t\text{dt}}{\sqrt{4\left( 1 - \sin^2{t} \right)}}\ }\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\frac{\frac{2}{3}\cos t\text{dt}}{2\sqrt{1 - \sin^2{t}}}\ }\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\frac{\frac{2}{3}\cos t\text{dt}}{2\sqrt{\cos^2{t}}}\ }\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\frac{\frac{2}{3}\cos\text{t }\text{dt}}{2\cos t}\ }\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\frac{\text{dt}}{3}\ }\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{t}{3} + C\ ;\ \ \ \text{substitute back for}\ t,\ \text{we have}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{3}\sin^{-1}\left( \frac{3x}{2} \right) + C\]
When we have an integrand that is a product of two functions of the same
variable such that we can we can write it in the form \(\int u\ dv\),
then we can have the following formula for solving the integral.
\[\int_{}^{}\text{u dv} = uv - \int_{}^{}\text{v du}\]
The function \(u\) is selected suitably so that \(\text{du}\) is less
complex.
Example
Integrate \(\int_{}^{}{x^{4}\ln x}\text{dx}\)
Solution
Let \(u = \ln x\) and \(dv = x^{4}\text{dx}\) so that
\(du = \frac{1}{x}\text{dx}\) and \(v = \frac{x^{5}}{5}\)
Then using integration by pats formula, we have
\[\int_{}^{}{x^{4}\ln x}\ dx = \frac{x^{5}}{5}\ln x - \int_{}^{}\frac{x^{5}}{5} \cdot \frac{1}{x}\text{dx}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{x^{5}}{5}\ln x - \frac{1}{5}\int_{}^{}x^{4}\text{dx}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{x^{5}}{5}\ln x - \frac{1}{5} \cdot \frac{x^{5}}{5} + C\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{x^{5}}{5}\ln x - \frac{x^{5}}{25} + C\]
When the integrand is a fraction whose power of numerator is more than
denominator, a long division is done so that the quotient is integrated
using the methods discussed above, The remaining part is solved using
partial fraction decomposition. In this case, the fraction is broken
down into as sum of smaller fractions whose denominator are irreducible
factors of the original fraction.
When a factor is repeated n times, then we have \(n\) fractions with
decominator associated with it were the factors are raised to \(1,2,3\)
up to \(n - 1\) and \(n\) respectively.
Example
\[\int_{}^{}\frac{- 2x^{2} + 3x + 2}{x^{3} + x^{2}}\]
Solution
\[\int_{}^{}\frac{- 2x^{2} + 3x + 2}{x^{3} + x^{2}} = \int_{}^{}\frac{- 2x^{2} + 3x + 2}{x^{2}(x + 1)}\]
Let
\[\frac{- 2x^{2} + 3x + 2}{x^{2}(x + 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x + 1}\]
Thus, we have
\[\frac{- 2x^{2} + 3x + 2}{x^{2}(x + 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x + 1} = \frac{\text{Ax}\left( x + 1 \right) + B\left( x + 1 \right) + Cx^{2}}{x^{2}(x + 1)}\]
Thus,
\[- 2x^{2} + 3x + 2 = Ax\left( x + 1 \right) + B\left( x + 1 \right) + Cx^{2}\]
We carry out substitution
When \(x = 0,\ 2 = B.1\), thus \(B = 2.\)
When \(x = - 1\), \(- 2 - 3 + 2 = C;\ \ C = - 3\)
When \(x = 1\),
\(3 = 2A + 2B + C = 2A + 2\left( 2 \right) + \left( - 3 \right)\). Thus,
\(3 = 2A + 1\) or \(A = 1.\)
Thus,
\[\frac{- 2x^{2} + 3x + 2}{x^{2}(x + 1)} = \frac{1}{x} + \frac{2}{x^{2}} - \frac{3}{x + 1}\]
\[\int_{}^{}\frac{- 2x^{2} + 3x + 2}{x^{3} + x^{2}} = \int_{}^{}{\left( \frac{1}{x} + \frac{2}{x^{2}} - \frac{3}{x + 1} \right)\text{dx}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\left( \frac{1}{x} + 2x^{- 2} - \frac{3}{x + 1} \right)\text{dx}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}\frac{\text{dx}}{x} + 2\int_{}^{}x^{- 2}dx - 3\int_{}^{}{\frac{\text{dx}}{x + 1}\ }\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ln\left| x \right| - 2x^{- 1} - 3\ln\left| x + 1 \right| + C\]
When an integral involves a quadratic equation, the equation is factored
by completing square method then solved using any of the appropriate
method discussed above.
Example
\[\int_{}^{}\frac{\left( 6x + 9 \right)\text{dx}}{4x^{2} + 12x + 13}\]
Solution
\[\int_{}^{}\frac{\left( 6x + 9 \right)\text{dx}}{4x^{2} + 12x + 13} = \int_{}^{}\frac{\left( 6x + 9 \right)\text{dx}}{\left( 4x^{2} + 12x + 9 \right) + 4} = \int_{}^{}\frac{3\left( 2x + 3 \right)\text{dx}}{\left( 2x + 3 \right)^{2} + 4}\]
Let \(u = 2x + 3\ \)then \(du = 2dx\ \)or \(\frac{\text{du}}{2} = dx\)
Upon substitution, we have
\[\int_{}^{}\frac{\left( 6x + 9 \right)\text{dx}}{4x^{2} + 12x + 13} = \int_{}^{}\frac{3u\frac{\text{du}}{2}}{u^{2} + 4}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{3}{2}\int_{}^{}\frac{\text{u du}}{u^{2} + 4}\]
Let \(m = u^{2} + 4\), then \(dm = 2u du\) or
\(\frac{\text{dm}}{2} = u du\)
\[\ \ \ \ \ = \frac{3}{2}\int_{}^{}\frac{\text{u du}}{u^{2} + 4} = \frac{3}{4}\int_{}^{}\frac{\text{dm}}{m}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{3}{4}\ln\left| m \right|\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{3}{4}\ln\left| u^{2} + 4 \right|\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{3}{4}\ln\left| {(2x + 3)}^{2} + 4 \right| + C\]
Example
\[\int_{}^{}\frac{\text{dx}}{\sqrt{x^{2} + 2x - 5}}\]
Solution
\[\int_{}^{}\frac{\text{dx}}{\sqrt{x^{2} + 2x - 5}} = \int_{}^{}\frac{\text{dx}}{\sqrt{x^{2} + 2x + 1 - 1 - 5}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}\frac{\text{dx}}{\sqrt{{(x + 1)}^{2} - 6}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}\frac{\text{dx}}{\sqrt{{(x + 1)}^{2} - \left( \sqrt{6} \right)^{2}}}\]
Let \(u = x + 1\ \), then \(du = dx\), then
\[\int_{}^{}\frac{\text{dx}}{\sqrt{{(x + 1)}^{2} - \left( \sqrt{6} \right)^{2}}} = \int_{}^{}\frac{\text{du}}{\sqrt{u^{2} - \left( \sqrt{6} \right)^{2}}}\]
Let \(u = \sqrt{6} \sec t\ \)then \(du = \sqrt{6} \sec t\tan t\text{dt}\) and
\(u^{2} = \operatorname{6}t\). Upon substitution, we have
\[\int_{}^{}\frac{\text{du}}{\sqrt{u^{2} - \left( \sqrt{6} \right)^{2}}} = \int_{}^{}\frac{\ \sqrt{6} \sec t\tan t\text{dt}\ }{\sqrt{6 \sec^{2} t - 6}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}\frac{\sqrt{6} \sec t\tan t\text{dt}\ }{\sqrt{6 (\sec^2{t}-1)}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}\frac{\sqrt{6} \sec t\tan t\text{dt}\ }{\sqrt{6\ \tan^2{t}}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}\frac{\sqrt{6} \sec t\tan t\text{dt}\ }{\sqrt{6}\tan t}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{}^{}{\sec t\text{dt}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ln\left| \sec t + \tan t \right| + C\ \]
When we are interested in determining an integral of of a function over
a given integral, then the integral gets limits within which it should
be evaluated. Such an integral is called a definite integral. The
definite integral over an interval bounded by numerical values gives a
number. Let \(g(x)\) and \(f(x)\) be functions such that
\[\frac{df(x)}{\text{dx}} = g(x)\]
Then the integral of \(g(x)\) with respect to \(x\) is
\[\int_{}^{}{g(x)}\text{dx}\]
and the definite integral of \(g(x)\) over a closed interval
\(\left\lbrack a,b \right\rbrack\) is
\[\int_{a}^{b}{g\left( x \right)}\ dx = f\left( b \right) - f\left( a \right)\text{.}\]
Example
Evaluate the definite integral
\[\int_{0}^{1}{\left( 9x^{2} + 5x + 1 \right)\text{dx}}\]
\[\int_{}^{}{\left( 9x^{2} + 5x + 1 \right)\text{dx}} = 3x^{3} + \frac{5x^{2}}{2} + x + C,\backslash n\]
\[\ \int_{0}^{1}{\left( 9x^{2} + 5x + 1 \right)\text{dx}} = \left\lbrack 3x^{3} + \frac{5x^{2}}{2} + x + C \right\rbrack_{0}^{1}\]
\[\ \ \ \ \ \ \ \ \ \ \ = \left( 3\left( 1 \right)^{3} + \frac{5{(1)}^{2}}{2} + 1 + C \right) - \left( 3{(0)}^{3} + \frac{5{(0)}^{2}}{2} + 0 + C \right)\]
\[\ \ \ \ \ \ \ \ \ \ \ = 3 + \frac{5}{2} + 1 + C - C\]
\[\ \ \ \ \ \ \ \ \ \ \ = \frac{13}{2}.\]
Here, you notice that the constant \(C\) is not part of the answer. It
is eliminated by subtraction of a similar constant. This happens in
definite integrals. For this reason, most definite integrals are
evaluated without including it in the procedure.
The definite integral is an indefinite integral where the limits are
included in the evaluation, as such, all the properties of indefinite
integrals applies here.
Let \(f(x)\) and \(g(x)\) be two functions integrable on a closed
interval \(\left\lbrack a,b \right\rbrack\) and \(k\) a constant. Then
\[\int_{a}^{b}{1\ dx} = b - a\]
\[\int_{a}^{b}{0\ dx} = 0\]
\[\int_{a}^{a}{f(x)\ dx} = 0\]
\[\int_{a}^{b}\text{k dx} = k(b - a)\]
\[\int_{a}^{b}{k\ f(x)dx} = k\int_{a}^{b}{\ f(x)dx}\]
\[\int_{a}^{b}{\left( f\left( x \right) \pm g\left( x \right) \right)\text{dx}} = \int_{a}^{b}{f\left( x \right)\text{dx}} \pm \int_{a}^{b}{g\left( x \right)\text{dx}} = b - a\]
\[\int_{a}^{b}{f\left( x \right)\text{dx}} = \int_{a}^{c}{f\left( x \right)\text{dx}} + \int_{c}^{b}{f\left( x \right)\text{dx}}\ \ ifa \leq c \leq b\]
Example
Find the integral of \(f\left( x \right) = \left\{ \begin{matrix}
x^{2} + 1,\ \ \& x < 3 \\
8x,\ \ \& x \geq 3 \\
\end{matrix} \right.\ \ \)on the interval
\(\left\lbrack 0,5 \right\rbrack.\)
Solution
\[\int_{0}^{5}{f\left( x \right)\text{dx}} = \int_{0}^{3}{f\left( x \right)\text{dx}} + \int_{3}^{5}{f\left( x \right)\text{dx}}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int_{0}^{3}{\left( x^{2} + 1 \right)\text{dx}} + \int_{3}^{5}{8x dx}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack \frac{x^{3}}{3} + 1 \right\rbrack_{0}^{3} + \left\lbrack {4x}^{2} \right\rbrack_{3}^{5}\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left( \frac{3^{3}}{3} + 3 \right) - \left( \frac{0^{3}}{3} + 0 \right) + \left( {4(5)}^{2} \right) - ({4(3)}^{2})\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 12 - 0 + 100 - 36\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 76.\]
a). Determination of area
Integrals are used to determine the area under a curve or between
curves. For instance, if \(f(x)\) in an integrable function, then the
intergral from point \(x = a\) to \(x = b\) is the area between the
curve, the \(x -\)axis and the lines, \(x = a\) and \(x = b.\)
\[\text{Area} = \int_{a}^{b}{f(x)}\text{dx}\]
b). Work done
Integral can also be used to determine the work done by a force. If a
point mass moves from one point to another in force field(electric
field, magnetic field), say \(\mathbf{F}\), then the work done
in moving the point is given by
\[\text{Area} = \int_{a}^{b}\text{F.dr}\]
Where \(\text{dr}\) is the differential representing the a general
position vector of a point along the given path is followed.
c). Velocity and Distance
Integration can also be used to determine the velocity or distance of a
moving object. When an acceleration of a body is given in terms of a
function of time \(t\), \(a(t)\) then the velocity function is given by
\[v\left( t \right) = \int_{a}^{b}{a(t)} \text{dt}\]
The speed is the absolute value \(\left| v\left( t \right) \right|.\)
On the other hand, if the velocity of a body if given in terms of the
function\(\ v(t)\), then the displacement from point \(a\) to \(b\) is
given by
\[s\left( t \right) = \int_{a}^{b}{v(t)}\text{dt}\]
The distance is given by the absolute value \(\left| s(t) \right|.\)
