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 Integrals

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