Integration is the inverse process of differentiation. The primary problem of Differential Calculus is "Given a function, to find its differential coefficient". But the primary problem of Integral Calculus is its inverse, (i. e), "Given the differential coefficient of a function, to find the function itself". There are several methods of integration in Integral Calculus. The methods include, direct anti derivative method, Method of substitution, Trigonometric Substitution, Special integrals, Integration by parts, Integration using Simpson's rule, Trapezoidal rule etc. The integrals which are not between two ordinates of the x -axis or abscissa of the y-axis are called indefinite integrals. We use an arbitrary constant "C" for indefinite integrals. We use the extreme points as the lower limits and the upper limits to evaluate the definite integrals. We shall see some of the solved examples of indefinite and definite integrals.

## Indefinite Integral Examples

### Solved Examples

Question 1: Evaluate $\int$ ( 5 x4 + 8 x3 - 6 x2 + 2x + 10 ) dx.
Solution:

We have $\int$ ( 5 x4 + 8 x3 - 6 x2 + 2x + 10 ) dx.

Using the formula $\int$ xn dx = $\frac{x^{n+1}}{n+1}$

We get, $\int$ ( 5x4 + 8 x3 - 6 x2 + 2x + 10 ) dx. =  5 $\frac{x^{5}}{5}$ + 8 $\frac{x^{4}}{4}$ - 6 $\frac{x^{3}}{3}$ + 2 $\frac{x^{2}}{2}$ + 10 x + c

#### $\int$ ( 5 x4 + 8 x3 - 6 x2 + 2x + 10 ) dx.   = x5 + 2 x4 - 2 x3 + x2 + 10 x + C

Question 2: $\int$ csc x ( csc x - cot x ) dx
Solution:

We have $\int$ csc x ( csc x - cot x ) dx = $\int$ ( csc2 x - csc x . cot x ) dx

= $\int$ csc2 x dx - $\int$ csc x. cot x dx

= - cot x - ( - csc x ) + C

= - cot x + csc x + C

$\int$ csc x ( csc x - cot x ) dx    = csc x - cot x + C

Question 3: $\int$ (3x + 4)7 . dx
Solution:

We have, I = $\int$ (3x + 4)7 . dx
Since the expansion of the binomial, (3x + 7)7 will contain more number of terms, we can integrate by substitution.

Substituting , u = 3x + 4

We get $\frac{du}{dx}$ = 3

=>      du = 3 . dx

=>      dx = $\frac{du}{3}$

Substituting in the given integral we get,

I = $\int$ (3x + 4)7 . dx

= $\int$ u7 . $\frac{du}{3}$

= $\frac{1}{3}$ $\int$ u7 . du

= $\frac{1}{3}$ . $\frac{u^{8}}{8}$ + C

$\int$ ( 3x + 4 )7 . dx   = $\frac{1}{24}$ . ( 3x + 4 )8 + C

Question 4: Evaluate $\int$ $\sqrt{1+cos2x}$ dx
Solution:

We have I = $\int$ $\sqrt{1+cos2x}$ dx

From the double angle formula in trigonometry, we have
cos 2x = 2 cos2 x - 1
=> cos 2x + 1 = 2 cos2 x
Substituting this in the given integral I, we get,
I = $\int$ $\sqrt{1+cos2x}$ dx

= $\int$ $\sqrt{2cos^{2}x}$  dx

= $\int$ $\sqrt{2}$ . cos x dx

= $\sqrt{2}$ $\int$  cos x dx

$\int$ $\sqrt{1+cos2x}$ dx      = $\sqrt{2}$  sin x + C

Question 5: Evaluate $\int$ $\frac{5}{3x-7}$ . dx
Solution:

We have I = $\int$ $\frac{5}{3x-7}$ . dx

Substituting u = 3x - 7,

we get $\frac{du}{dx}$ = 3

=>                      du = 3 . dx

=>                      dx = $\frac{du}{3}$

Substituting in the given integral, we get,

I = $\int$ $\frac{5}{3x-7}$ . dx

= $\int$ $\frac{5}{u}$ $\frac{du}{3}$

= $\frac{5}{3}$ . $\int$ $\frac{1}{u}$. du

= $\frac{5}{3}$ . ln | u | + C

$\int$ $\frac{5}{3x-7}$ . dx     = $\frac{5}{3}$ . ln | ( 3x - 7 ) | + C

## Definite Integral Examples

Fundamental Theorem of Integral calculus: Let F(x) be the primitive or anti derivative of a continuous function f(*x) defined on [a, b].
( i. e ) $\frac{d}{dx}$ [ F(x)] = f(x ).
Then the definite integral of f ( x ) over [ a, b ] is denoted by $\int_{a}^{b}$ f(x) dx = F(b) - F(a)
The numbers a and b are called the limits of integration, 'a' is called the lower limit and 'b' is called the upper limit.
The interval [a,b] is called the interval of integration.
Algorithm to evaluate definite integral.
Step 1: Find the indefinite integral $\int$ f(x) dx. Let it be F(x). There is no need to keep the constant of integration.
Step 2: Evaluate F(b) and F(a) .
Step 3: Calculate F(b) - F(a).
The number obtained in Step 3 is the value of the definite integral $\int_{a}^{b}$ f(x) dx.

### Solved Examples

Question 1: Evaluate $\int_{1}^{2}$ x2  dx
Solution:

We have I = $\int_{1}^{2}$ x2  dx

We know that $\int$ x2 . dx = $\frac{x^{3}}{3}$

Therefore, we have F(x) = $\frac{x^{3}}{3}$

F(2) = $\frac{2^{3}}{3}$

= $\frac{8}{3}$ and

F(1) = $\frac{1^{3}}{3}$

= $\frac{1}{3}$

According to Step 3, we have F(2) - F(1) = $\frac{8}{3}$ - $\frac{1}{3}$

$\int_{1}^{2}$ x2  dx     = $\frac{7}{3}$

Question 2: $\int_{0}^{\pi/2}$ cos3 x . dx
Solution:

We have, $\int_{0}^{\pi/2}$ cos3 x . dx = $\int_{0}^{\pi/2}$ $\frac{cos3x+3cosx}{4}$ dx    [ since cos 3x = 4 cos3 x - 3 cos x ]

= $\frac{1}{4}$ [ $\frac{sin3x}{3}$ + 3 sin x ]$_{0}^{\pi/2}$

= $\frac{1}{4}$ [ ( $\frac{sin3\pi/2}{3}$ + 3 sin $\pi/2$ ) - $\frac{sin0}{3}$ + 3 sin 0 ) ]

= $\frac{1}{4}$ [ ( - $\frac{1}{3}$ + 3 ) - ( 0 + 0 ) ]

$\int_{0}^{\pi/2}$ cos3 x . dx      = $\frac{2}{3}$

Question 3: $\int_{1}^{2}$ $\frac{3x}{9x^{2}-1}$ . dx
Solution:

We have,              $\int_{1}^{2}$ $\frac{3x}{9x^{2}-1}$ . dx

Substituting, u = 9 x2 - 1,

we get, $\frac{du}{dx}$ = 18 x

=>    du = 18 x dx

=> x dx = $\frac{du}{18}$

Therefore,            I = $\int$ $\frac{3x}{9x^{2}-1}$ . dx

= 3 $\int$ $\frac{xdx}{9x^{2}-1}$

= 3 $\int$ $\frac{1}{u}$ $\frac{du}{18}$

= $\frac{3}{18}$ $\int$ $\frac{du}{u}$

= $\frac{1}{6}$ ln | u |

= $\frac{1}{6}$ ln | 9 x - 1 |

$\int_{1}^{2}$ $\frac{3x}{9x^{2}-1}$ . dx =   $\frac{1}{6}$ [ln | 9 x - 1 | ]$_{1}^{2}$

= $\frac{1}{6}$ [ [ln | 9 (2) - 1 | - [ln | 9 (1) - 1 | ]

= $\frac{1}{6}$ [ ln | 35| - ln | 8 | ]

$\int_{1}^{2}$ $\frac{3x}{9x^{2}-1}$ . dx = $\frac{1}{6}$ ln | 35/8 |

Question 4: $\int$ $\frac{1}{a^{2}+x^{2}}$
Solution:

We have

I = $\int_{0}^{a}$ $\frac{1}{a^{2}+x^{2}}$ dx

Substituting  x = a tan$\theta$

we get    $\frac{dx}{d\theta}$ = a sec2 $\theta$

=> dx = a sec2 $\theta$ . d$\theta$
Therefore, substituting in the given integral, we get,

I = $\int$ $\frac{1}{a^{2}+x^{2}}$ dx

= $\int$ $\frac{1}{a^{2}+a^{2}tan^{2}\theta}$ a sec2 $\theta$ d$\theta$

= $\int$ $\frac{a}{a^{2}(1+tan^{2}\theta)}$ a sec2 $\theta$ d$\theta$

= $\frac{1}{a}$ $\int$ $\frac{1}{sec^{2}\theta}$ sec2 $\theta$ d$\theta$

= $\frac{1}{a}$ . $\int$ d$\theta$

=  $\frac{1}{a}$  $\theta$

$\int_{0}^{a}$ $\frac{1}{a^{2}+x^{2}}$ dx      = $\frac{1}{a}$ [ arctan ($\frac{x}{a}$)]$_{0}^{a}$

= $\frac{1}{a}$ [arctan ( 1 ) - arc tan ( 0 )]

= $\frac{1}{a}$$\frac{\pi}{4} - 0 \int_{0}^{a} \frac{1}{a^{2}+x^{2}} dx = \frac{1}{a}$$\frac{\pi}{4}$

Question 5: $\int_{-1}^{1}$ f(x) dx, where f (x ) =  1 - 2x, x $\le$ 0
= 1 + 2x , x$\ge$ 0.
Solution:

We have,
$\int_{-1}^{1}$ f(x) dx = $\int_{-1}^{0}$ f(x) dx + $\int_{0}^{1}$ f(x) dx

= $\int_{-1}^{0}$ ( 1 - 2 x ) dx + $\int_{0}^{1}$ ( 1 + 2 x ) dx

= ( x - x2 )$_{-1}^{0}$  + ( x + x2 )$_{0}^{1}$

= [ 0 - ( -1 - 1 ) ] + [( 1 + 1 - ( 0 ) ]

= 2 + 2 = 4

$\int_{-1}^{1}$ f(x) dx  = 4