Written By Keerthi Kulkarni
Last Modified 24-12-2024

Standard Forms: Integration, Forms, Rules, Formulas

Standard Forms in Math: Integration is a method that brings disparate parts together to form a whole. We find a function whose differential is known in integral calculus. As a result, integration is the polar opposite of differentiation. Integration gives the area under the curve of \(f(x)\). It is also known as the entire class of antiderivatives and is read as an indefinite integral of \(f\) with respect to \(x\).

Integrals are used in various fields, including Engineering, where engineers use integrals to determine the shape of a building. In Physics, for example, it is used to find the centre of gravity. Also, three-dimensional models are demonstrated in the field of graphical representation.

Standard Forms: Definition of Integration

Finding the area under a curve is known as integration. This is achieved by drawing as many small rectangles as possible to cover the area under the curve. Then, the areas are added together. The sum approaches a limit equal to the region of a function under the curve.

The process of obtaining an antiderivative of a function is known as integration. It is the definite integration if a function is integrable and integral over the domain is finite, with the bounds defined.

Standard Form of Indefinite Integrals

The integration of any function with no bounds is known as an indefinite integral. There are no upper or lower bounds for an indefinite integral. The antiderivative of a function is the indefinite integral, which is the inverse of differentiation.

If the derivative of a function \(f(x)\) is represented by \({f^\prime }(x)\) the integration of the consequence \({f^\prime }(x)\) returns the original function \(f(x)\)

Let us discuss the standard forms of indefinite integral by using the expression mentioned.

\(\frac{{df(x)}}{{dx}} = {f^\prime }(x)\)

\(\int {{f^\prime }} (x)dx = f(x) + C\)

Standard Form of Definite Integral

The area under a curve in a graph can be calculated using the definite integral. It has a start point and an endpoint within which the area under a curve is determined. Hence, we can say that it has limits.

The area under a curve that lies between two set boundaries is called a definite integral. To determine the area of the curve \(f(x)\) with respect to the \(x-\)axis, as the limit points \([a,b]\) the analogous expression for the definite integral is \(\int_a^b f (x)dx\)

Where \(a\) is the lower limit and \(b\) is the upper limit.

Standard Integration Formulas

The following six sets of integration formulas have been described in broad strokes. Essentially, integration is a method of bringing dissimilar parts together to form a whole. An operation that is the inverse of differentiation is called integration. As a result, the basic integration formula is \(\int {{f^\prime }} (x)dx = f(x) + C\)

where \(C\) is an arbitrary constant.

Standard Form of Integration by Parts

Use the integration by parts formula or partial integration to evaluate the integral when the function given is a product of two other functions. When partial integration is used, the integration formula is given by:

\(\int f (x) \cdot g(x) = f(x)\int {\left( {\int g (x)dx \cdot {f^\prime }(x)} \right)} dx + C\)

Standard Form of Integration by Substitution

Use the integration formula for substitution when a function is of another function.

If \(I = \int f (x)dx\), where \(x = g(t)\) so that \(\frac{{dx}}{{dt}} = {g^\prime }(t)\), then we write \(dx = {g^\prime }(t)\). We can write \(I = \int f (x)dx = \int f (g(t)){g^\prime }(t)dt\)

Standard Form of Integration by Partial Fractions

Utilise integration by partial fractions to get the integral of \(\frac{{P(x)}}{{Q(x)}}\) that is an improper fraction, where \(P(x)\) and \(Q(x)\) are polynomials, and the degree of \(P(x)\) is less than the degree of \(Q(x)\) Then,

\(\frac{{P(x)}}{{Q(x)}} = T(x) + \frac{{{P_1}(x)}}{{Q(x)}}\)

where \(T(x)\) is a polynomial in \(x\) and \(\frac{{{P_1}(x)}}{{Q(x)}}\) is a valid rational function and is used to split the fraction. If \(A, B\) and \(C\) are real numbers, the following categories of simpler partial fractions are connected with different forms of rational functions.

Form of Rational Fractions	Form of Partial Fractions
\(\frac{{px + q}}{{(x – a)(x – b)}}\)	\(\frac{A}{{x – a}} + \frac{B}{{x – b}}\)
\(\frac{{px + q}}{{{{(x – a)}^n}}}\)	\(\frac{{{A_1}}}{{x – a}} + \frac{{{A_2}}}{{{{(x – a)}^2}}} + \cdots \frac{{{A_n}}}{{{{(x – a)}^n}}}\)
\(\frac{{p{x^2} + qx + r}}{{a{x^2} + bx + c}}\)	\(\frac{{Ax + B}}{{a{x^2} + b + c}}\)
\(\frac{{p{x^2} + qx + r}}{{{{\left( {a{x^2} + bx + c} \right)}^n}}}\)	\(\frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + \cdots \frac{{{A_n}x + {B_n}}}{{{{\left( {a{x^2} + bx + c} \right)}^n}}}\)
\(\frac{{p{x^2} + qx + r}}{{(x – a)(x – b)(x – c)}}\)	\(\frac{A}{{x – a}} + \frac{B}{{x – b}} + \frac{C}{{x – c}}\)
\(\frac{{p{x^2} + qx + r}}{{{x^2} + bx + c}}\)	\(\frac{A}{{x – a}} + \frac{{Bx + C}}{{{x^2} + bx + c}}\)

Basic Integration Formulas

Generalised results are produced using the fundamental theorems of integrals, which are known as integration formulas in indefinite integration.

Integration	Result
\(\int {{x^n}} dx\)	\(\frac{{{x^{n + 1}}}}{{n + 1}} + C\)
\(\int 1 \,dx\)	\(x + C\)
\(\int {{e^x}} dx\)	\({e^x} + C\)
\(\int {\frac{1}{x}} dx\)	\(\log \|x\| + C\)
\(\int {{a^x}} dx\)	\(\frac{{{a^x}}}{{\log a}} + C\)
\(\int {{e^x}} \left[ {f(x) + {f^\prime }(x)} \right]dx\)	\({e^x}f(x) + C\)

Practice 10th CBSE Exam Questions

Standard Formulas of Integration Using Trigonometric Functions

Here are a few key integration formulas to remember for quick and accurate calculations. When dealing with trigonometric functions, we simplify them and rewrite them as integrable functions.

Function	Integral Value
\(\int {\cos } \,x\,dx\)	\(\sin\,x + C\)
\(\int {\sin } \,x\,dx\)	\( – \cos \,x + C\)
\(\int {{{\sec }^2}} \,x\,dx\)	\(\tan x + C\)
\(\int {{{{\mathop{\rm cosec}\nolimits} }^2}} \,x\,dx\)	\( – \cot x + C\)
\(\int {\sec } \,x \cdot \tan \,x\,dx\)	\(\sec x + C\)
\(\int {{\mathop{\rm cosec}\nolimits} } \,x \cdot \cot x\,dx\)	\( – {\mathop{\rm cosec}\nolimits} x + C\)
\(\int {\tan } \,x\,dx\)	\(\log \|\sec x\| + C\)
\(\int {\cot } \,x\,dx\)	\(\log \|\sin x\| + C\)
\(\int {\sec } \,x\,dx\)	\(\log \|\sec x + \tan x\| + C\)
\(\int {{\mathop{\rm cosec}\nolimits} } \,x\,dx\)	\(\log \|{\mathop{\rm cosec}\nolimits} \,x – \cot x\| + C\)

Standard Integration Formulas Involving Inverse Trigonometric Functions

A list of inverse trigonometric functions is provided in the below table:

Function	Integration Value
\(\int {\frac{1}{{\sqrt {1 – {x^2}} }}} dx\)	\({\sin ^{ – 1}}x + C\)
\(\int {\frac{1}{{1 – {x^2}}}} dx\)	\( – {\cos ^{ – 1}}x + C\)
\(\int {\frac{1}{{1 + {x^2}}}} dx\)	\({\tan ^{ – 1}}x + C\)
\(\int {\frac{{ – 1}}{{1 – {x^2}}}} dx\)	\( – {\cot ^{ – 1}}x + C\)
\(\int {\frac{1}{{x\sqrt {{x^2} – 1} }}} dx\)	\({\sec ^{ – 1}}x + C\)
\(\int {\frac{{ – 1}}{{x\sqrt {{x^2} – 1} }}} dx\)	\( – {{\mathop{\rm cosec}\nolimits} ^{ – 1}}x + C\)

Standard Integration Formulas Involving Trigonometric Functions of Power “n”

The list of trigonometric formulas of trigonometric functions with power \(n\) is tabulated below:

Function	Integration Value
\(\int {{{\sin }^n}} \,x\,dx\)	\( – \frac{1}{n}{\sin ^{n – 1}}x \cdot \cos \,x + \frac{{n – 1}}{n}\int {{{\sin }^{n – 2}}} \,x\,dx\)
\(\int {{{\cos }^n}} \,x\,dx\)	\(\frac{1}{n}{\cos ^{n – 1}}x \cdot \sin x + \frac{{n – 1}}{n}\int {{{\cos }^{n – 2}}} \,x\,dx\)
\(\int {{{\tan }^n}} \,x\,dx\)	\(\frac{1}{{n – 1}}{\tan ^{n – 1}}x + \int {{{\tan }^{n – 2}}} \,x\,dx\)
\(\int {{{\sec }^n}} \,x\,dx\)	\(\frac{1}{{n – 1}}{\sec ^{n – 2}}x \cdot \tan x + \frac{{n – 2}}{{n – 1}}\int {{{\sec }^{n – 2}}} \,x\,dx\)
\(\int {{{{\mathop{\rm cosec}\nolimits} }^n}} \,x\,dx\)	\(\frac{{ – 1}}{{n – 1}}{{\mathop{\rm cosec}\nolimits} ^{n – 2}}x \cdot \cot x + \frac{{n – 2}}{{n – 1}}\int {{{{\mathop{\rm cosec}\nolimits} }^{n – 2}}} \,x\,dx\)

Solved Examples – Standard Forms

Below are a few solved examples that can help in getting a better idea:

Q.1. Find the value of \(\int {\frac{{9x + 25}}{{{{(x + 3)}^2}}}} dx\)
Ans: Given: \(\frac{{9x + 25}}{{{{(x + 3)}^2}}}\) is a rational function

Using the partial fraction decomposition, we have \(\frac{{9x + 25}}{{{{(x + 3)}^2}}} = \frac{A}{{x + 3}} + \frac{B}{{{{(x + 3)}^2}}}\)

\(\frac{{9x + 25}}{{{{(x + 3)}^2}}} = \frac{{[A(x + 3) + B]}}{{{{(x + 3)}^2}}}\)

Equating the numerator, we get

\(9x + 25 = A(x + 3) + B\)

Solving for \(B\) when \(x=-3\), we get \(B=-2\)

Solving for \(A\) when \(x=0\), we get \(A=9\)

Thus, the partial fraction is decomposed as \(\frac{9}{{x + 3}} – \frac{2}{{{{(x + 3)}^2}}}\)

\(\int {\frac{{9x + 25}}{{{{(x + 3)}^2}}}} \cdot dx = \int {\left[ {\frac{9}{{x + 3}}} \right]} dx – \int {\frac{2}{{{{(x + 3)}^2}}}} \cdot dx\)

\( = 9\,l{\rm{n}}(x + 3) – \frac{2}{{x + 3}} + C\)

Q.2. Simplify and find the value of \(\int {{\mathop{\rm cosec}\nolimits} } \,x({\mathop{\rm cosec}\nolimits}\,x – \cot x)dx\) using the integration formula.
Ans: Given: \(\int {{\mathop{\rm cosec}\nolimits} } \,x({\mathop{\rm cosec}\nolimits} x – \cot x) \cdot dx\)

\( = \int {{{{\mathop{\rm cosec}\nolimits} }^2}} x \cdot dx – \int {\cot } \,x \cdot {\mathop{\rm cosec}\nolimits}\,x \cdot dx\)

Using the basic trigonometric integration formulas,

\( = – \cot x – ( – {\mathop{\rm cosec}\nolimits}\,x) + C\)

\( = {\mathop{\rm cosec}\nolimits}\,x – \cot x + C\)

Q.3. Use the integration formula and find the value of \(\int {\frac{{5 + 4\cos x}}{{{{\sin }^2}x}}} \cdot dx\)
Ans: Given, \(\int {\frac{{5 + 4\cos x}}{{{{\sin }^2}x}}} \cdot dx\)

\( = \int {\frac{5}{{{{\sin }^2}x}}} \cdot dx + \int {\frac{{4\cos x}}{{{{\sin }^2}x}}} \cdot dx\)

\( = \int 5\,{{\mathop{\rm cosec}\nolimits} ^2}x \cdot dx + \int 4 \cot x \cdot {\mathop{\rm cosec}\nolimits}\,x \cdot dx\)

\( = 5\int {{{{\mathop{\rm cosec}\nolimits} }^2}} x \cdot dx + 4\int {\cot } \,x \cdot {\mathop{\rm cosec}\nolimits}\,x \cdot dx\)

Using the trigonometric integration formula, we get

\( = 5( – \cot x) + 4( – {\mathop{\rm cosec}\nolimits}\,x)\)

\( = – 5\cot x – 4\,{\mathop{\rm cosec}\nolimits}\,x + C\)

Q.4. Find the value of \(\int {\frac{{2x}}{{1 + {x^2}}}} dx\).
Ans: Consider \(1 + {x^2} = t\),

So, \(2\,x\,dx = dt\)

So, the given integration becomes

\(\int {\frac{1}{t}} dt\)

By using the formula

\(\int {\frac{1}{t}} dt = \log |t| + C\)

Substituting the value of \(t\),

Thus, \(\log \left| {1 + {x^2}} \right| + C\)

Hence, the value of \(\int {\frac{{2x}}{{1 + {x^2}}}} dx\) is \(\log \left| {1 + {x^2}} \right| + C\)

Q.5. Find the value of \(\int {{{\tan }^2}} (2x – 3)dx\)
Ans: Consider \(2x – 3 = t\)

So, \(2\,dx = dt\)

Thus, the integration becomes \(\frac{1}{2}\int {{{\tan }^2}} \,t\,dt\)

The value of \(\int {{{\tan }^2}} \,t\,dt\) can be calculated by the formula

\(\int {{{\tan }^n}} \,x\,dx = \frac{1}{{n – 1}}{\tan ^{n – 1}}x + \int {{{\tan }^{n – 2}}} \,x\,dx\)

Thus,

\( = \frac{1}{2}\left[ {\frac{1}{{2 – 1}}{{\tan }^{2 – 1}}t + \int {{{\tan }^{2 – 2}}\,} t\,dt} \right]\)

\( = \frac{1}{2}\left\{ {\tan t + \int t \,dt} \right\}\)

\( = \frac{1}{2}\tan t + \frac{1}{2}t + C\)

Substitute the value of \(t\), we will get

\( = \frac{1}{2}\tan (2x – 3) + \frac{1}{2}(2x – 3) + C\)

\( = \frac{1}{2}\tan (2x – 3) + x + C\)

Hence, the value of \(\int {{{\tan }^2}} (2x – 3)dx\) is \(\frac{1}{2}\tan (2x – 3) + x + C\).

Summary

The process of finding the area of the region under the curve is known as integration. The process of finding the antiderivative of the function is known as integration of the function. The standard form of indefinite integral is \(\int f(x)dx\), whereas the standard form of definite integral is \(\int_a^b f(x)dx\), where \(a,\,b\) are lower and upper limits.

There are various standard forms of representing and finding the integration of the functions such as by partial fractions, by parts and by substitution etc.

There are standard formulas are associated with integration which involves trigonometry and inverse trigonometry functions helps in solving the integral calculus. Integration has wide range of applications in the fields of Mathematics, Physics, Engineering and Architecture and Medicine.

FAQs on Standard Form in Math

Students might be having many questions with respect to the Standard Forms. Here are a few commonly asked questions and answers:

Q.1. What is standard form?
Ans: Standard form is the method of representing a particular element in the most common manner.

Q.2. What is the standard form of indefinite integral?
Ans: The standard form of indefinite integral is given by \(\int f (x)dx = F(x) + C\), where \(F(x)\) is the antiderivative of the function \(f(x)\) and \(C\) is the arbitrary constant.

Q.3. What is the standard formula for the integration by parts?
Ans: The formula for the integration by parts is given by \(\int f (x) \cdot g(x) = f(x) \cdot \int {\left( {\int g (x)dx \cdot {f^\prime }(x)} \right)} dx + C\).

Q.4. What is the standard form of definite integral?
Ans: The standard form of definite integral is given by \(\int_a^b f (x)dx = F(b) – F(a) + C\), where \(a, b\) are the limits or boundaries of the function.

Q.5. Mention any two standard integration formulas that uses trigonometric functions.
Ans: The standard integration formulas using trigonometric functions is given below:
\(\int {\cos } \,x\,dx = \sin x + C\)
\(\int {\sin } \,x\,dx = – \cos x + C\)

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