What is Integration, Its Methods, Properties and Formulae?

Integration is the inverse process of differentiation that collects infinitely many infinitesimal contributions of a function. It is used when the derivative of a function is given and we are asked for its original function.

Function Cos x is the derived function of Sin x, whereas Sin x will be the anti-derivative of Cos x. IIT JAM and JEE Main often ask questions related to integration. Let us understand integration with an example:

 

Let us now look at some basic integration formulas used:

In this chapter, students are familiarised with the concept of integration which is important for CBSE board exams. Students gain an insight into how a function can be found when its derivative is given and learn to derive the area and volume of graphs within specific limits. This chapter is also about definite and indefinite integration and the techniques employed to obtain these integrals.The students can invest some effort, learn the formulas, and conduct a thorough study to make it one of the most interesting topics for examination point of view. Integration is a subsection that comes under the section “calculus”. It carries nominal marks in class X. It carries 5 marks in class XI. It carries 35 marks, in class XII. 

There are various methods for finding integration in any given function. NEET exam aspirant must learn about these methods to ensure that they are able to solve integration problems of all types:

• By substitutions: This method is used when the variable is substituted by another ideal variable to ease the integration process. It is used when part of the integrand is derivative of another part. Let us consider an example: 

∫2x·cos(x²) dx → set u=x², 
 du=2x dx, 
it becomes ∫cos u du.

• By parts: It is used whenever you have the product of 2 functions. One of them will become simpler when you will differentiate it. The formula used here is ∫u dv = u·v – ∫v du. Let us consider an example to understand this method.

∫x·eˣ dx

let u=x (du=dx), dv=eˣ dx (v=eˣ),

so result = x eˣ – ∫eˣ dx.

• By using partial fractions: This method of integration is used on a fracton which has a denominator that factors into simpler pieces. Let use take an example. Say, we will split (Ax+B)/(x²–1) into A/(x–1) + B/(x+1). After this, we will integrate all pieces so that they become  ln|x–1|, ln|x+1|).
• Trigonometric Substitution: This integration method is used on √(a²–x²), √(a²+x²), or √(x²–a²). For example:

 ∫dx/√(a²–x²)
 set x=a sinθ, 
then dx=a cosθ dθ and √(a²–x²)=a cosθ.

Integration with trigonometric properties can easily solve complex integrals. For example, expressions such as  $\sqrt{a^2-x^2},\sqrt{a^2+x^2}or\sqrt{x^2-a^2}$ will be easily converted into plain angles once you set $x=a\sin \left(\theta \right)$, $x=a\tan \left(\theta \right)$ or $x=a\sec \left(\theta \right)$. Through trigonometric properties, half-angle formulae turn them into first-order cosines of multiples. Here is a list of trigonometric identities:

• Direct integrals of trig functions. $\int \sin \left(x\right)dx=-\cos \left(x\right)+C$ , $\int \cos \left(x\right)dx=\sin \left(x\right)+C$.

• Power‐reduction for even powers. ${\sin }^2\left(x\right)=\frac{1-\cos \left(2x\right)}{2},{\cos }^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}$ . Example: $\int {\sin }^2\left(x\right)dx=\int \frac{1-\cos \left(2x\right)}{2}dx=\frac{x}{2}-\frac{\sin \left(2x\right)}{4}+C$ .

• Odd powers ⇒ peel off one factor + substitute. Example: $\int {\sin }^5\left(x\right)dx=\int ((1-\cos \left(x\right){)}^2\sin \left(x\right))dx(u=\backslash cos\; x;\; du=-\backslash sin\; x\backslash ,dx)$ .

• Products ⇒ sum‐to‐product. $\sin \left(x\right)\cos \left(x\right)=\frac{1}{2}\sin \left(2x\right)$ . Example: $\int \sin \left(x\right)\cos \left(x\right)dx=\frac{1}{2}\int \sin \left(2x\right)dx=-\frac{1}{4}\cos \left(2x\right)+C$ .

• Secant and tangent tricks. $ddx\tan \left(x\right)={\sec }^2\left(x\right)$ , $ddx\sec \left(x\right)=\sec \left(x\right)\tan \left(x\right)$ . Examples: $\int {\sec }^2\left(x\right)dx=\tan \left(x\right)+C$ , $\int \sec \left(x\right)\tan \left(x\right)dx=\sec \left(x\right)+C$ .

• Trig substitutions for algebraic integrals. To handle $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$, set $x=a\sin \left(\theta \right)$, $x=a\tan \left(\theta \right)$, or $x=a\sec \left(\theta \right)$.

1. Find the anti-derivatives of the following functions.

(i) cos3x

Solution. We have ∫ cos3x = (1/3)sin3x + C, where C is the constant of integration.

(ii) sin2x - 4e3x

Solution. We have ∫sin2x dx -  ∫ 4e3x dx =(-cos2x/2+ C1) - (4/3 e3x +C2 )

= -cos2x/2 - 4/3 e3x + C

Here C =  C1 - C2, which is the constant of integration in this case.

(iii) (2- 3 sinx)/cos²x

Solution. We have  ∫ 2/cos² x dx - 3sinx/cos² x dx

= ∫ 2sec²x dx - ∫3secx tanx dx

= 2tanx - 3secx + C

C is again the constant of integration.

2. Solve ∫ 3ax/(b2 +c2x2) dx

Solution. Let us take v = b2 +c2x2, then

dv = 2c2x dx

Thus, ∫ 3ax/(b2 +c2x2) dx

= (3ax/2c2x)∫dv/v

= (3a/2c2)∫dv/v

= (3a/2c2) log |b2 +c2x2| + C

3. Write the antiderivatives of the function: 3x2+4x3

Solution. ∫3x2+4x3 dx = 3(x3/3) + 4(x4/4)

= x3 + x4

Thus, the antiderivative of 3x2+4x3 = x3 + x4

4. Integrate the function 2x sin(x2+ 1) with respect to X, using substitution method

Solution. Given function: 2x sin(x²+ 1)

Using the substitution method, we get

x² + 1 = t, so that 2x dx = dt.

= ∫ 2x sin ( x² +1) dx = ∫ sint dt

= – cos t + C

= – cos (x² + 1) + C

Where C is an arbitrary constant

Therefore – cos (x² + 1) + C