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Indefinite and Definite Integration

In Calculus, integration and differentiation are two most important concepts. Integration originated during the course of finding the area of a plane figure whereas differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Integration is the reverse of differentiation. It is also called as the antiderivative. In this section, aspirants will learn about the indefinite and definite Integration list of important formulas, how to use integral properties to solve integration problems, integration methods and many more.

Indefinite Integration

The indefinite integral is defined as a function that will describe an area under the function’s curve from an undefined point to another arbitrary point. Read more.

Standard formulas for Indefinite Integration

  1. \(\begin{array}{l}\int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}}+c\end{array} \)
  2. \(\begin{array}{l}\int{{{a}^{x}}dx}=\frac{{{a}^{x}}}{\ell n\,a}+c\end{array} \)
  3. \(\begin{array}{l}\int{{{e}^{x}}}dx={{e}^{x}}+c\end{array} \)
  4. \(\begin{array}{l}\int{\frac{1}{x}}dx=\ell n\,x+c\end{array} \)
  5. \(\begin{array}{l}\int{\sin x\,dx=-\cos x+c}\end{array} \)
  6. \(\begin{array}{l}\int{\cos x\,dx=\sin x+c}\end{array} \)
  7. \(\begin{array}{l}\int{{{\sec }^{2}}x\,\,dx=\tan x+c}\end{array} \)
  8. \(\begin{array}{l}\int{\cos e{{c}^{2}}x\,\,dx=-\cot x\,+c}\end{array} \)
  9. \(\begin{array}{l}\int{\sec x\tan xdx=\sec x+c}\end{array} \)
  10. \(\begin{array}{l}\int{\cos ec\,\,x\,\,}\cot x\,dx=-\cos ec\,x+c\end{array} \)
  11. \(\begin{array}{l}\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx\begin{matrix} = \\ = \\ \end{matrix}}\left\{ \begin{matrix} {{\sin }^{-1}}x+c \\ -{{\cos }^{-1}}x+c \\ \end{matrix} \right.\end{array} \)
  12. \(\begin{array}{l}\int{\frac{1}{1+{{x}^{2}}}}dx\begin{matrix} = \\ = \\ \end{matrix}\left\{ \begin{matrix} {{\tan }^{-1}}x+c \\ -{{\cot }^{-1}}x+c \\ \end{matrix} \right.\end{array} \)
    .
  13. \(\begin{array}{l}\frac{1}{x\sqrt{{{x}^{2}}-1}}=\left\{ \begin{matrix} {{\sec }^{-1}}x+c \\ -\cos e{{c}^{-1}}x+c \\ \end{matrix} \right.\end{array} \)

⇒ Also Read – Introduction to Integration

Methods of Integration

  1. Substitution Method
  2. Integration by partial fractions
  3. By parts
  4. Euler substitution
  5. Reduction method

Properties of Indefinite Integration

  1. \(\begin{array}{l}k\int{f\left( x \right)dx=k\int{f\left( x \right)}}\end{array} \)
  2. \(\begin{array}{l}\int{\left( f\left( x \right)+g\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx}}}\end{array} \)

Below are some Illustrations based on integration properties and substitution method:

Illustration 1: Solve 

\(\begin{array}{l}\int{\frac{{{\left( 1+x \right)}^{3}}}{\sqrt[3]{{{x}^{2}}}}}dx\end{array} \)

Solution: 

\(\begin{array}{l}\int{\frac{{{\left( 1+x \right)}^{3}}}{\sqrt[3]{{{x}^{2}}}}}dx=\int{\frac{1+3x+3{{x}^{2}}+{{x}^{3}}}{{{x}^{2/3}}}}dx\end{array} \)
\(\begin{array}{l}=\int({{{x}^{-2/3}}+3{{x}^{1/3}}+3{{x}^{4/3}}}+{{x}^{7/3}})dx\end{array} \)
\(\begin{array}{l}=3{{x}^{1/3}}+\frac{9}{4}{{x}^{4/3}}+\frac{9}{7}{{x}^{7/3}}+\frac{3}{10}{{x}^{10/3}}+c\end{array} \)

Illustration 2: Solve 

\(\begin{array}{l}\int{{\sec }^{2}}x\cos e{{c}^{2}}x\,\,dx\end{array} \)

Solution:

\(\begin{array}{l}\int{{{\sec }^{2}}x\cos e{{c}^{2}}x\,\,dx=\int{\frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x\,{{\cos }^{2}}x}}dx}\end{array} \)
\(\begin{array}{l}=\int{{{\sec }^{2}}x+\cos e{{c}^{2}}x\,\,dx}\end{array} \)
\(\begin{array}{l}=\tan x-\cot x+c\end{array} \)

Illustration 3.

\(\begin{array}{l}\int{2x\,\,\sin \left( {{x}^{2}} \right)dx}\end{array} \)

Solution: Put

\(\begin{array}{l}{{x}^{2}}=t\end{array} \)

then,

\(\begin{array}{l}2x\,\,dx=dt\end{array} \)
,

\(\begin{array}{l}\int{\sin t\,\,dt=-\cos t+c}\end{array} \)
,

\(\begin{array}{l}=-\cos {{x}^{2}}+c\end{array} \)

Illustration 4.

\(\begin{array}{l}\int{{{\sin }^{3}}x{{\cos }^{5}}x\,\,dx}\end{array} \)

Solution:

\(\begin{array}{l}\int{{{\sin }^{2}}x{{\cos }^{5}}x\sin x\,\,dx}\end{array} \)
\(\begin{array}{l}=\int{\left( 1-{{\cos }^{2}}x \right)}{{\cos }^{5}}x\,\,\sin x\,\,dx\end{array} \)

put

\(\begin{array}{l}\cos x=t\end{array} \)

\(\begin{array}{l}\,\,\,-\sin x\,dx=dt\end{array} \)

so, new equation is:

\(\begin{array}{l}\int{\left( 1-{{t}^{2}} \right)}{{t}^{5}}\left( -dt \right)\end{array} \)
\(\begin{array}{l}=\int{{{t}^{5}}-{{t}^{7}}}dt\end{array} \)
\(\begin{array}{l}=-\left( \frac{{{t}^{6}}}{6}-\frac{{{t}^{8}}}{8} \right)\end{array} \)

Resubstituting the value, we get

\(\begin{array}{l}=-\frac{{{\cos }^{6}}x}{6}+\frac{{{\cos }^{8}}x}{8}+c\end{array} \)

Important formulae set for Indefinite Integration

  1. \(\begin{array}{l}\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \frac{x}{a} \right)+c\end{array} \)
  2. \(\begin{array}{l}\int{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|}\end{array} \)
  3. \(\begin{array}{l}\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|}\end{array} \)
  4. \(\begin{array}{l}\int{\frac{1}{{{a}^{2}}+{{x}^{2}}}}dx=\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right).\end{array} \)
  5. \(\begin{array}{l}\int{\frac{1}{{{a}^{2}}-{{x}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{a+x}{a-x} \right|.\end{array} \)
  6. \(\begin{array}{l}\int{\frac{1}{{{x}^{2}}-{{a}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{x-a}{x+a} \right|.\end{array} \)
  7. \(\begin{array}{l}\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{a} \right).\end{array} \)
  8. \(\begin{array}{l}\int{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\ell n\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|\end{array} \)
  9. \(\begin{array}{l}\int{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}+\frac{{{a}^{2}}}{2}\ell n\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|\end{array} \)

 Types of substitutions:

Type I

\(\begin{array}{l}\int{{{\sin }^{m}}x{{\cos }^{n}}x\,\,dx}\end{array} \)

Rule:

  1. If m, n one of them odd then substitute for even power.
  2. If both odd then substitute either of them.
  3. If both even use trigonometric identities.

Type II: Substitution of Trigonometric functions.

  1. \(\begin{array}{l}\sqrt{{{a}^{2}}-{{x}^{2}}}\rightarrow{{}}x=a\sin \theta\end{array} \)
  2. \(\begin{array}{l}\sqrt{{{x}^{2}}+{{a}^{2}}}\rightarrow{{}}x=a\tan \theta\end{array} \)
  3. \(\begin{array}{l}\sqrt{{{x}^{2}}-{{a}^{2}}}\rightarrow{{}}x=a\sec \theta\end{array} \)
  4. \(\begin{array}{l}\sqrt{\frac{x-a}{b-x}}\rightarrow{{}}x=a{{\cos }^{2}}\theta +b{{\sin }^{2}}\theta\end{array} \)
  5. \(\begin{array}{l}\sqrt{\frac{x-a}{x-b}}\rightarrow{{}}x=a{{\sec }^{2}}\theta -b\end{array} \)

Type III:

\(\begin{array}{l}\int{\frac{px+q}{a{{x}^{2}}+bx+c}}dx\And \int{\frac{px+q}{\sqrt{a{{x}^{2}}+bx+c}}}\,dx\And \int{\left( px+q \right)}\sqrt{a{{x}^{2}}+bx+c}\,dx\end{array} \)

We express

\(\begin{array}{l}px+q\end{array} \)
as
\(\begin{array}{l}m{{\left( a{{x}^{2}}+bx+c \right)}^{1}}+n\end{array} \)
then this gets changed to standard integral.

Biquadratic Substitutions

1.

\(\begin{array}{l}I=\int{f\left( x+\frac{1}{x} \right)}\left( 1-\frac{1}{{{x}^{2}}} \right)dx\end{array} \)

Put

\(\begin{array}{l}x+\frac{1}{x}=t\end{array} \)

2.

\(\begin{array}{l}I=\int{f\left( x-\frac{1}{x} \right)}\left( 1+\frac{1}{{{x}^{2}}} \right)dx\end{array} \)

Put

\(\begin{array}{l}\left( x-\frac{1}{x} \right)=t\end{array} \)

Solved Problems

Problem 1: Solve 

\(\begin{array}{l}\int{\frac{dx}{\sqrt{{{x}^{2}}+2x+2}}}\end{array} \)

Solution:

\(\begin{array}{l}\int{\frac{dx}{\sqrt{{{x}^{2}}+2x+2}}}=\int{\frac{dx}{\sqrt{{{\left( x+1 \right)}^{2}}+1}}}\end{array} \)
\(\begin{array}{l}=\log \left| \left( x+1 \right)+\sqrt{{{\left( x+1 \right)}^{2}}+1} \right|.\end{array} \)

Problem 2: Solve

\(\begin{array}{l}\int{\sqrt{5{{x}^{2}}+2x+3}}\,\,dx\end{array} \)

Solution:

\(\begin{array}{l}\int{\sqrt{5{{x}^{2}}+2x+3}}\,\,dx\end{array} \)

here

\(\begin{array}{l}5{{x}^{2}}+2x+3=5\left[ {{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}} \right]\end{array} \)
,

\(\begin{array}{l}\int{\sqrt{5{{x}^{2}}+2x+3}\,dx}=\sqrt{5}\int{\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}}}}\end{array} \)
,

\(\begin{array}{l}=\sqrt{5}\frac{\left( x+\frac{1}{5} \right)}{2}\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{15}} \right)}^{2}}}+\frac{14}{25.2}\log \left| \left( x+\frac{1}{5} \right)+\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+\frac{14}{15}} \right|\end{array} \)

Problem 3: Evaluate 

\(\begin{array}{l}I=\int{\frac{x+1}{{{x}^{2}}+3x+4}}dx\end{array} \)

Solution:

\(\begin{array}{l}I=\int{\frac{x+1}{{{x}^{2}}+3x+4}}dx\end{array} \)
,

\(\begin{array}{l}x+1=a{{\left( {{x}^{2}}+3x+4 \right)}^{1}}+b\end{array} \)
,

\(\begin{array}{l}=2ax+\left( 3a+b \right)\end{array} \)
[Using Type III (mentioned above)]

\(\begin{array}{l}a=\frac{1}{2}\,\,\,\,b=\frac{-1}{2}\end{array} \)

\(\begin{array}{l}I=\int{\frac{1}{2}}\frac{{{\left( {{x}^{2}}+3x+4 \right)}^{1}}}{{{x}^{2}}+3x+4}-\frac{1}{2}\left( \frac{1}{{{x}^{2}}+3x+4} \right)\end{array} \)
,

\(\begin{array}{l}=\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right)-\frac{1}{2}\int{\frac{1}{{{\left( x+\frac{3}{2} \right)}^{2}}+\frac{7}{4}}}\end{array} \)
,

\(\begin{array}{l}=\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right)-\frac{1}{2}\frac{1}{\sqrt{\frac{7}{4}}}{{\tan }^{-1}}\frac{\left( x+\frac{3}{2} \right)}{\sqrt{\frac{7}{4}}}\end{array} \)
,

\(\begin{array}{l}=\frac{1}{2}\log \left| {{x}^{2}}+3x+4 \right|-\frac{1}{\sqrt{7}}{{\tan }^{-1}}\left( \frac{2x+3}{\sqrt{7}} \right)\end{array} \)

Problem 4: Solve

\(\begin{array}{l}\int{\frac{1+{{x}^{2}}}{1+{{x}^{4}}}}\,dx\end{array} \)

Solution:

\(\begin{array}{l}\int{\frac{1+{{x}^{2}}}{1+{{x}^{4}}}}\,dx=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{\left( \frac{1}{{{x}^{2}}}+{{x}^{2}} \right)}}\,dx\end{array} \)
,

\(\begin{array}{l}=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)dx}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}}\end{array} \)
,

\(\begin{array}{l}=\int{\frac{d\left( x-\frac{1}{x} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}}\end{array} \)
,

\(\begin{array}{l}=\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\left( x-\frac{1}{x} \right)}{\sqrt{2}} \right)\end{array} \)

Integration by Partial Fraction

Integrals of rational functions can be evaluated by splitting them into partial fractions.

\(\begin{array}{l}\frac{f\left( x \right)}{g\left( x \right)}\end{array} \)
where
\(\begin{array}{l}f\And g\end{array} \)
are polynomial is called as rational function.

If the degree of

\(\begin{array}{l}f<\end{array} \)
, the degree of g it is called a proper fraction.

Otherwise, it is an improper fraction.

Then

\(\begin{array}{l}\frac{f\left( x \right)}{g\left( x \right)}=h\left( x \right)+\frac{d\left( x \right)}{g\left( x \right)}\end{array} \)
, degree of d < degree of g.

Cases:

1. When

\(\begin{array}{l}g\left( x \right)\end{array} \)
is expressed as product of non-repeating linear factors.

\(\begin{array}{l}g\left( x \right)=\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)….\left( x-{{a}_{n}} \right)\end{array} \)
then

\(\begin{array}{l}\frac{f}{g}=\frac{{{A}_{1}}}{x-{{a}_{1}}}+\frac{{{A}_{2}}}{x-{{a}_{2}}}+….\frac{{{A}_{n}}}{x-{{a}_{n}}}.\end{array} \)

2. Some factors of g are repeating then

\(\begin{array}{l}g\left( x \right)={{\left( x-a \right)}^{k}}\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)…\end{array} \)
\(\begin{array}{l}\frac{f}{g}=\frac{{{A}_{1}}}{\left( x-a \right)}+\frac{{{A}_{2}}}{{{\left( x-a \right)}^{2}}}+…..\frac{{{A}_{k}}}{{{\left( x-a \right)}^{n}}}+…\end{array} \)

3.

\(\begin{array}{l}g\left( x \right)\end{array} \)
has a quadratic term then
\(\begin{array}{l}\frac{f}{g}=\frac{Ax+B}{a{{x}^{2}}+bx+c}\end{array} \)

Where A & B are constants determined by comparing coefficients.

Let us understand with the help of examples:

Example 1: Solve

\(\begin{array}{l}\int{\frac{dx}{\left( x+1 \right)\left( x-2 \right)}}\end{array} \)

Solution:

\(\begin{array}{l}\int{\frac{dx}{\left( x+1 \right)\left( x-2 \right)}}=\int{\frac{a}{x+1}+\frac{b}{x-2}}\,dx\end{array} \)
,

\(\begin{array}{l}=\int{\frac{-\frac{1}{3}}{x+1}}+\int{\frac{\frac{1}{2}}{x-2}}\,dx\end{array} \)
,

\(\begin{array}{l}=-\frac{1}{3}\ell n\left| x+1 \right|+\frac{1}{2}\ell n\left| x-2 \right|\end{array} \)

Example 2: Solve

\(\begin{array}{l}\int{\frac{2{{x}^{2}}-1}{\left( x-1 \right){{\left( x+1 \right)}^{2}}}}dx\end{array} \)

Solution:

\(\begin{array}{l}\int{\frac{2{{x}^{2}}-1}{\left( x-1 \right){{\left( x+1 \right)}^{2}}}dx=\int{\frac{A}{x-1}+\frac{B}{\left( x+1 \right)}}+\frac{C}{{{\left( x+1 \right)}^{2}}}}\end{array} \)
,

\(\begin{array}{l}2{{x}^{2}}-1=A{{\left( x+1 \right)}^{2}}+B\left( x+1 \right)\left( x+1 \right)+C\left( x-1 \right)\end{array} \)

Put

\(\begin{array}{l}x=1\Rightarrow A=\frac{1}{2}\end{array} \)
,

\(\begin{array}{l}x=-1\Rightarrow \,\,C=\frac{-1}{2}\end{array} \)
,

\(\begin{array}{l}B=\frac{3}{2}\end{array} \)
,

\(\begin{array}{l}I=\int{\frac{\frac{1}{2}}{\left( x-1 \right)}}dx+\frac{3}{2}\int{\frac{dx}{\left( x+1 \right)}}+\left( \frac{-1}{2} \right)\int{\frac{dx}{{{\left( x-2 \right)}^{2}}}}\end{array} \)
,

\(\begin{array}{l}=\frac{1}{2}\ell n\left| x-1 \right|+\frac{3}{2}\left( ln\left| x+1 \right| \right)+\frac{1}{2\left( x+2 \right)}+c\end{array} \)
,

Example 3: Solve 

\(\begin{array}{l}\int{\frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}}\end{array} \)

Solution:

\(\begin{array}{l}\int{\frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}}=\frac{A}{x+2}+\frac{Bx+C}{{{x}^{2}}+1}\end{array} \)

Comparing by

\(\begin{array}{l}x=-2\Rightarrow \,\,A=\frac{1}{5}\end{array} \)
and by
\(\begin{array}{l}{{x}^{2}}\end{array} \)
coeff.
\(\begin{array}{l}B=\frac{-1}{2}\end{array} \)
and
\(\begin{array}{l}C=\frac{2}{5}\end{array} \)
,

on substituting the values and integrating we have

\(\begin{array}{l}I=\frac{1}{5}\ell n\left| x+2 \right|-\frac{1}{10}\ell n\left( {{x}^{2}}+1 \right)+\frac{2}{5}{{\tan }^{-1}}\left( x \right)\end{array} \)

Integration of Trigonometric Functions

Type 1:

\(\begin{array}{l}I=\int{{{\sin }^{m}}x{{\cos }^{n}}xdx}\end{array} \)

1. If m –odd put

\(\begin{array}{l}\cos x=t\end{array} \)

2. If n odd put

\(\begin{array}{l}\sin x=t\end{array} \)

3. If m, n rationales then put

\(\begin{array}{l}\tan x=t\end{array} \)

4. If both even then use reduction method

Let us understand with the help of an example:

\(\begin{array}{l}\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}}\end{array} \)

Where

\(\begin{array}{l}t=\sin x\end{array} \)
\(\begin{array}{l}=\int{{{t}^{-6}}-{{t}^{-4}}dt}\end{array} \)
\(\begin{array}{l}=-\frac{1}{5si{{n}^{5}}x}+\frac{1}{3{{\sin }^{3}}x}+c\end{array} \)

Type 2:

\(\begin{array}{l}\int{\frac{dx}{a\cos x+b\sin x+c}}\end{array} \)

Put

\(\begin{array}{l}t=\tan \left( \frac{x}{2} \right)\end{array} \)

Illustration:

\(\begin{array}{l}\int{\frac{dx}{2+\sin x}}\end{array} \)
\(\begin{array}{l}\Rightarrow t=\tan \left( \frac{x}{2} \right)\end{array} \)
\(\begin{array}{l}dx=\frac{2dt}{1+{{t}^{2}}}\end{array} \)
\(\begin{array}{l}=\int{\frac{\frac{2dt}{1+{{t}^{2}}}}{2+\frac{2t}{1+{{t}^{2}}}}}\end{array} \)
\(\begin{array}{l}\Rightarrow \int{\frac{dt}{{{t}^{2}}+t+1}}\end{array} \)
\(\begin{array}{l}=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right)\end{array} \)
\(\begin{array}{l}=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\tan \frac{x}{2}+1}{\sqrt{3}} \right)+c\end{array} \)

Type 3

\(\begin{array}{l}\int{\frac{dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}},\int{\frac{dx}{a+b{{\sin }^{2}}x}}\end{array} \)
,

\(\begin{array}{l}\int{\frac{1}{a+b{{\cos }^{2}}x}}\,dx,\int{\frac{1}{{{\left( a\sin x+b\cos x \right)}^{2}}}}\,dx\end{array} \)
or
\(\begin{array}{l}\int{\frac{1}{a+b{{\sin }^{2}}x+{{\cos }^{2}}x}}\,dx\end{array} \)

Rule:

Divide both numerator & denominator by

\(\begin{array}{l}{{\cos }^{2}}x\end{array} \)

Illustration:

\(\begin{array}{l}\int{\frac{1}{3-4{{\sin }^{2}}x}}\,dx=\int{\frac{\frac{1}{{{\cos }^{2}}x}}{\frac{3}{{{\cos }^{2}}x}-\frac{4{{\sin }^{2}}x}{{{\cos }^{2}}x}}}\end{array} \)
\(\begin{array}{l}=\int{\frac{{{\sec }^{2}}x}{3{{\sec }^{2}}x-4{{\tan }^{2}}x}}\,dx\end{array} \)
\(\begin{array}{l}=\int{\frac{dt}{3\left( 1+{{t}^{2}} \right)-4{{t}^{2}}}}\end{array} \)
(Since, by
\(\begin{array}{l}\tan x=t\,\,\,{{\sec }^{2}}x\,dx=dt\end{array} \)
)

\(\begin{array}{l}=\int{\frac{dt}{3-{{t}^{2}}}}\end{array} \)
\(\begin{array}{l}=\int{\frac{1}{{{\left( \sqrt{3} \right)}^{2}}-{{t}^{2}}}}\end{array} \)
\(\begin{array}{l}=\frac{1}{2\sqrt{3}}\ell n\left| \frac{\sqrt{3}+t}{\sqrt{3}-t} \right|\end{array} \)
\(\begin{array}{l}=\frac{1}{2\sqrt{3}}\log \left| \frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x} \right|\end{array} \)

Indefinite Integration – Video Lesson

Indefinite Integration - Video Lesson

Indefinite-Integration Problems

Indefinite-Integration Problems

Definite Integrals

Definite Integrals

Choose n divide [a, b] in n parts of width

\(\begin{array}{l}h=\frac{b-a}{n}=\end{array} \)
. partition of interval putting (n – 1) parts in between

\(\begin{array}{l}\int\limits_{a}^{b}{f\left( x \right)}dx=\underset{h\to 0}{\mathop{\lim }}\,h\,\sum\limits_{n=0}^{n-1}{f\left( a+rh \right)}\end{array} \)

Definition of Definite Integral as limit of a solution

\(\begin{array}{l}=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+…..+f\left( a+\left( n-1 \right)h \right) \right]\end{array} \)
,

\(\begin{array}{l}nh=b-a\end{array} \)

Integrals using limits

Evaluate as Limit of a Sum

\(\begin{array}{l}I=\int\limits_{a}^{b}{x\,\,dx\Rightarrow f\left( x \right)=x\,\,\,so\,h=\frac{b-a}{n}}\end{array} \)
,

\(\begin{array}{l}f\left( a \right)=a\end{array} \)
,

\(\begin{array}{l}f\left( a+h \right)=a+h\end{array} \)
,

\(\begin{array}{l}I=h\left[ a+\left( a+h \right)+….\left( a+\left( n-1 \right)h \right) \right]\end{array} \)
look for nh

\(\begin{array}{l}=h\left[ \left( na \right)+h+2h+3h….\left( n-1 \right)h \right]\end{array} \)
,

\(\begin{array}{l}=hna+\frac{{{h}^{2}}n\left( n-1 \right)}{2}\end{array} \)
,

\(\begin{array}{l}=\left( b-a \right)a+\frac{{{h}^{2}}{{n}^{2}}}{2}-\frac{{{\left( b-a \right)}^{2}}}{2}=\frac{{{b}^{2}}-{{a}^{2}}}{2}\end{array} \)
,

\(\begin{array}{l}\sum\limits_{r=1}^{n}{n}=\frac{n\left( n+1 \right)}{2}\end{array} \)
,

\(\begin{array}{l}\sum\limits_{r=1}^{n}{{{n}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\end{array} \)
,

\(\begin{array}{l}\sum\limits_{r=1}^{n}{{{n}^{3}}}={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}\end{array} \)
,

GP …

\(\begin{array}{l}a+ar+a{{r}^{2}}….a{{r}^{n-1}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\end{array} \)
,

\(\begin{array}{l}\sin \alpha +\sin \left( \alpha +\beta \right)+\sin \left( \alpha +2\beta \right)….\sin \left( \alpha +\left( n-1 \right)\beta \right)\end{array} \)
,

\(\begin{array}{l}=\frac{\sin \left( n\frac{\beta }{2} \right)}{\sin \left( \frac{\beta }{2} \right)}\sin \left[ \frac{\alpha +\alpha +\left( n-1 \right)\beta }{2} \right]\end{array} \)
,

\(\begin{array}{l}=\frac{\sin \left( n\frac{diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\sin \left( \frac{1st+last}{2} \right)\end{array} \)
,

\(\begin{array}{l}=\frac{\sin \left( \frac{n\,diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\cos \left( \frac{1st+last}{2} \right)\end{array} \)
,

\(\begin{array}{l}1-\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}-\frac{1}{{{6}^{2}}}=\frac{{{\pi }^{2}}}{12}\end{array} \)
,

\(\begin{array}{l}1+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}….=\frac{{{\pi }^{2}}}{12}\end{array} \)
,

\(\begin{array}{l}\int\limits_{a}^{b}{{{e}^{x}}dx=\underset{n\to \infty }{\mathop{\lim }}\,}h\left[ {{e}^{a}}+{{e}^{a+h}}+{{e}^{a+2h}}…{{e}^{a+\left( n-1 \right)h}} \right]\end{array} \)
,

\(\begin{array}{l}=h{{e}^{a}}\left[ 1+{{e}^{n}}+{{e}^{2h}}+…. \right]\end{array} \)
,

\(\begin{array}{l}={{e}^{a}}\left[ {{e}^{nh}}-1 \right]\frac{h}{\left( {{e}^{n}}-1 \right)}\end{array} \)
,

\(\begin{array}{l}={{e}^{b}}-{{e}^{a}}\end{array} \)
,

\(\begin{array}{l}\int\limits_{a}^{b}{\sin x\,dx=\underset{n\to \infty }{\mathop{\lim }}\,}h\left[ \sin a+\sin \left( a+n \right)+…\sin \left( a+\left( n-1 \right)h \right) \right]\end{array} \)
,

\(\begin{array}{l}=\underset{n\to \infty }{\mathop{\lim }}\,h\frac{\sin \left( \frac{nh}{2} \right)}{\sin \left( \frac{h}{2} \right)}\left[ \sin \left( \frac{2a+\left( n-1 \right)h}{2} \right) \right]\end{array} \)
,

\(\begin{array}{l}=\frac{\sin \left( \frac{b-a}{2} \right)}{\left( \frac{1}{h} \right)\sin \left( \frac{h}{2} \right)}\left[ \sin \left( a+\frac{nh}{2}-\frac{h}{2} \right) \right]=2\sin \left( \frac{b-a}{2} \right)\sin \left( \frac{a+b}{2} \right)\end{array} \)
,

\(\begin{array}{l}=\cos \left( \frac{b-a-\left( a+b \right)}{2} \right)-\cos \left( \frac{\left( b-a \right)+\left( a+b \right)}{2} \right)\end{array} \)
\(\begin{array}{l}=\cos a-\cos b\end{array} \)
,

\(\begin{array}{l}\int\limits_{0}^{2}{\left( {{x}^{2}}+1 \right)dx=\frac{14}{3}}\end{array} \)
\(\begin{array}{l}\int\limits_{0}^{2}{\left( 3{{x}^{2}}+2x+1 \right)dx=3}\end{array} \)
,

\(\begin{array}{l}\int\limits_{0}^{\frac{\pi }{2}}{\cos x\,dx=1}\end{array} \)
or
\(\begin{array}{l}\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x\,dx=h\left[ \frac{1-\cos 2x}{2}+\frac{1-{{\cos }^{\left( 2n+2n \right)}}}{2}…… \right]}\end{array} \)
,

\(\begin{array}{l}=\frac{1}{2}\left( b-a \right)-\frac{1}{4}\left( \sin 2b-\sin 2n \right)\end{array} \)

Limits using Definite Integration

\(\begin{array}{l}\int\limits_{a}^{b}{f\left( x \right)dx=\underset{h\to 0}{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{f\left( a+rh \right)}}\end{array} \)

Special case a = 0 and b = 1

\(\begin{array}{l}\int\limits_{0}^{1}{f\left( x \right)\,dx=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{n}}\sum\limits_{r=1}^{n}{f\left( \frac{r}{n} \right)}\end{array} \)
\(\begin{array}{l}\int\limits_{0}^{k}{f\left( x \right)\,dx=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{n}}\sum\limits_{r=1}^{kn}{f\left( \frac{r}{n} \right)}\end{array} \)
\(\begin{array}{l}\sum{\rightarrow{Multiple\,\,of\,n}}\int{{}}\end{array} \)
\(\begin{array}{l}\frac{1}{n}\to dx\,\,\,\,\frac{r}{n}\Rightarrow x\end{array} \)

 

How to solve integration with the help of limits:

\(\begin{array}{l}\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+…\frac{1}{nb} \right]\end{array} \)
\(\begin{array}{l}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=0}^{\left( b-a \right)}{\frac{1}{na+r}}\end{array} \)
\(\begin{array}{l}=\sum\limits_{n\to \infty }{\frac{1}{n}\sum\limits_{r=0}^{\left( b-a \right)n}{\frac{1}{a+\frac{r}{n}}}}=\int\limits_{0}^{b-a}{\frac{dx}{a+x}}=\left[ \log \left( a+x \right) \right]_{a}^{b-a}=\log \left( \frac{b}{a} \right)\end{array} \)

Applications of Integrals Videos

Definite Integral & Area Under the Curve – Important Topics

Definite Integral & Area Under the Curve - Important Topics

Definite Integral & Area Under the Curve – Important Questions

Definite Integral & Area Under the Curve - Important Questions

Frequently Asked Questions

What do you mean by Integration?

Integration is the method of finding the antiderivative of a function.

What are the rules of Integration?

There are many rules of integration namely the power rule, the sum and difference rules, the exponential rule, the reciprocal rule, the constant rule, the substitution rule, and the rule of integration by parts etc.

Why is Integration important?

We utilise integration in mathematics to find areas, volumes, displacement, etc. Integral calculus originated from the concept of integration in calculus.