Methods of Integration: Definition, Formula, Problem & Class 12 Maths Notes

Using the example below, you can see which type of integrals can be used for algebraic integration.  

The idea here is that we need to simplify a given function whose integral needs to be found. That we could use to show it as the algebraic sum of constants and functions. These are below.  

• (ax + b)ⁿ       
• $\frac{1}{1+x^2}$                
• $\frac{1}{\sqrt[]{1-x^2}}$            
• $\frac{1}{x\sqrt[]{x^2-1}}$

Here is an example. 

Example: Find $\int \frac{dx}{\sqrt[]{x}+\sqrt[]{x+1}}$

Solution: $\int \frac{dx}{\sqrt[]{x}+\sqrt[]{x+1}}$

= $\int \frac{(\sqrt[]{x}-\sqrt[]{x+1})}{(\sqrt[]{x}+\sqrt[]{x+1)}(\sqrt[]{x}-\sqrt[]{x+1})}dx$

$\int \frac{(\sqrt[]{x}-\sqrt[]{x+1})}{x-(x+1)}dx$ = $\int \frac{(\sqrt[]{x}-\sqrt[]{x+1})}{-1}dx$  

= $\int (\sqrt[]{x+1}-\sqrt[]{x})dx$ = $\int {(x+2)}^{\frac{1}{2}}dx$ – $\int x^{\frac{1}{2}}dx$

= $\frac{{(x+2)}^{\frac{3}{2}}}{\frac{3}{2}}$  – $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$ + c = $\frac{2}{3}{(x+2)}^{\frac{3}{2}}$ – $\frac{2}{3}{x}^{\frac{3}{2}}$  + c

In this, we need to express the given integrand as the algebraic sum of the functions. We need to know these. 

(i) sin kθ                           (ii) cos kθ                         (iii) sec² kθ

(iv) cosec² kθ                  (v) sec kθ tan kθ            (vi) cosec kθ cot kθ

For this, use the following formulae, whichever are applicable:

(i) sin²x =    $\frac{1-cos2x}{2}$         (ii) cos³x = $\frac{3cosx+cos3x}{4}$ (iii) sin³x = $\frac{3sinx-sin3x}{4}$

(iv) cos²x = $\frac{1+cos2x}{2}$        (v) tan²x = sec²x – 1     (vi) cot²x = cosec²x – 1

(vii) 2sin A sin B = cos (A – B) – cos (A + B)

(viii) 2cos A cos B = cos (A – B) + cos (A + B)

(ix) 2sin A cos B = sin (A + B) + sin (A – B)

(x) 2cos A sin B = sin (A + B) – sin (A – B)

 

Example: Find $\int \frac{{sec}^{2}x}{{cosec}^{2}x}dx$

Solution: $\int \frac{{sec}^{2}x}{{cosec}^{2}x}dx$  = $\int \frac{1}{{cos}^{2}x}\times {sin}^{2}xdx$ = $\int {tan}^{2}xdx$

= $\int {(sec}^{2}x-1)dx$  = $\int {sec}^{2}xdx$  – $\int 1dx$

= tanx – x + c

You should be using these when required. 

(i) $\int a^{x}dx$ =      $\frac{a^x}{\log a}$                      (ii) $e^{alog}{e}^b$ = $b^a$            (iii) $e^{{log}_{e}a}$ = a

 

Example: Evaluate $\int a^{3x+3}dx$ , a>0

Solution: $\int a^{x}dx$  = $\frac{a^x}{{log}_{e}a}$

$\therefore$ $\int a^{3x}dx$ = $\int {{(a}^3)}^{x}dx$ = $\frac{{{(a}^3)}^x}{{log}^{a^3}}$ + c = $\frac{a^{3x}}{3loga}$  + c

  $\therefore$ $\int a^{3x+3}dx$ = $a^3\int a^{3x}dx$  = $a^3$ .  + $\frac{a^{3x}}{3loga}$ c

P(x). ${(ax+b)}^n$ or $\frac{P\left(x\right)}{{(ax+b)}^n}$

Where P(x) is a polynomial in x and n is a positive rational number.

Working Rule:

Put z = ax + b

Example: Evaluate $\int (5x+3)\sqrt[]{2x-1}dx$

Solution: Here, P(x) = (5x + 3) and $\sqrt[]{2x-1}$  is of form ${(ax+b)}^n$ .

Let z = 2x – 1, then $\frac{dz}{dx}$  = 2, $\therefore$  dx = $\frac{dz}{2}$

Again, z = 2x – 1 $\therefore$ x = $\frac{z+1}{2}$

Now, $\int (5x+3)\sqrt[]{2x-1}dx$  = $\int \left[5\left(\frac{z+1}{2}\right)+3\sqrt[]{z}\right]\frac{dz}{2}$

= $\int \frac{5z+11}{4}\sqrt[]{z}dz$  = $\int \left(\frac{5}{4}{z}^{\frac{3}{4}}+\frac{11}{4}\sqrt[]{z}\right)dz$

= $\frac{5}{4}$ $\int \left(z^{\frac{3}{4}}\right)dz$  + $\frac{11}{4}\int \sqrt[]{z}$ dz

= $\frac{5}{4}$ . $\frac{z^{\frac{5}{2}}}{\frac{5}{2}}$  + $\frac{11}{4}$   . $\frac{z^{\frac{3}{2}}}{\frac{3}{2}}$  + c = $\frac{1}{2}$ $z^{\frac{5}{2}}$    + $\frac{11}{6}$ $z^{\frac{3}{2}}$    + c

= $\frac{z^{\frac{3}{2}}}{2}$  (z + $\frac{11}{3}$ ) + c = $\frac{z^{\frac{3}{2}}}{2}$  ( $\frac{3z+11}{3}$  ) + c

= $\frac{{(2x-1)}^{\frac{3}{2}}}{6}.$   $\left[3\left(2x-1\right)+11\right]$  + c

= $\frac{{(2x-1)}^{\frac{3}{2}}}{6}.$  (6x + 8) + c = $\frac{{(2x-1)}^{\frac{3}{2}}.(3x+4)}{3}$ + c.

To find integrals of the product of functions, we need to use integration by parts. 
You can remember that we can find indefinite integrals by using the same formula. 

Formula for Integration by Parts

If u and v are two functions of x, then

$\int uvdx$  = u $\int vdx$ – $\int \left(\frac{du}{dx}\int vdx\right)$ dx                              ……… $\int xcosxdx$ ……………(1)

It's important to note that

• Integral of the product of two functions is equal to the first function $\times$  
• Integral of second function is Integral of [ d.c. of first function $\times$  Integral of second function]

To prove it, see the example below. 

If u and w be two functions of x, then

$\frac{d}{dx}$ (uw) = u $\frac{dw}{dx}$  + w $\frac{du}{dx}$

Integrating both sides in relation to x, we get

uw = $\int \left(u\frac{dw}{dx}\right)dx$  + $\int \left(w\frac{du}{dx}\right)dx$

$\therefore$ $\int u\frac{dw}{dx}dx$ = uw – $\int \left(w\frac{du}{dx}\right)dx$                                     ……………………(2)

Let $\frac{dw}{dx}$  = v, then w = $\int vdx$                                                 ……………………(3)   

$\therefore$ From (2), $\int uvdx$ = u $\int vdx$  – $\int \left(\frac{du}{dx}\int vdx\right)$ dx

Example: Evaluate $\int xcosxdx$          

Solution: Let u = x and v = cosx, then $\frac{du}{dx}$ = 1 and $\int cosxdx$ = sinx

Now, $\int uvdx$  = u $\int vdx$  – $\int \left(\frac{du}{dx}\int vdx\right)$ dx

$\therefore$ $\int xcosxdx$ = xsinx – $\int \left(1.sinx\right)dx$

 = xsinx – (– cosx) + c = xsinx + cosx + c

From the above example that we just did, we need to know that the integral of functions of the form uv may be found by the method of integration by parts. This will happen when u and v are two different functions and their derivatives do not connect them.
But if derivatives connect them, try using the substitution method in the beginning. Once done, go on to use integration by parts.    

Some functions may not be products of two functions. We can still find their integrals using integration by parts. We can do that by taking 1 as the second function.

Some common examples of this condition would be logarithmic and inverse trigonometric functions.

Important Topics:

NCERT Class 11 notes	CBSE NCERT Class 11 Physics
Class 11 Chemistry notes	Class 11 Maths notes

We need to know that we can evaluate integrals by writing down the partial fraction of the integrand if they meet two conditions.

• Integrand is of the form $\frac{P\left(x\right)}{Q\left(x\right)}$ . In this, both P(x) and Q(x) are polynomials in x.
• Also remember that Q(x) has only linear and quadratic factors.

Before you write, know that the partial fraction of P(x)/Q(x) has the highest power of x in the numerator P(x). That should be made less than the highest power of x in Q(x). You can do that by dividing P(x) by Q(x).

Also Read: NCERT Solution for Class 11 & 12