`(sin(sqrt(x)+a)*e^sqrt(x))/sqrt(x)`

Result


Error
Variables
Answer
Steps
Plot

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.
abc
Shows the alphabet.
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 \(\left| s(t) \right|.\)