Integration by Substitution: Overview, Questions, Preparation

Integration is a method under calculus, opposite of differentiation, where we find the integrals of functions. Integration by substitution is one of the methods to solve integrals. This method is also called u-substitution.

The integration of a function f(x) is given by F(x), where F(x) is obtained by substitution. It is represented by:
∫f(x)dx = F(x) + C

The anti-derivatives of basic functions are known to us as integrals. But this integration technique is limited to basic functions and determines the integrals of various functions; so different integration methods are used. Among these methods of integration, the following is integration by substitution.

Here, any given integral is transformed into a simple form of integral by substituting others’ independent variable. The point to be kept in mind is that the substituted variable’s differentiation should also exist in the equation. Only then the equation is suitable to be solved using Integration by Substitution.

Example:

Integrate 2x/(1 + x2)

Let 2x = t, (1 + x2) = dt,

∫t.dt = t2/2 + C = (2x)2/2 + C = 4x2/2 + C = 2x2 + C

Details of the Topic

This topic is an essential part of the integration topic and is introduced in Class 12 to solve complex equations to obtain a simpler solution. Calculus problems carry 35 marks in board exams and give a deeper understanding of the integration and differentiation topic. Knowing the integrals are mandatory to solve the problems faster. Other methods of integration are essential, but this by far has maximum importance in exams. Students will also understand differentiation better in this topic, and they will learn to recognise when and where to apply this method on seeing the question.

1. Solve ∫sin(x³).3x².dx with respect x

Solution. Substituting x³ = t --(2)

Differentiation of the above equation will give:

3x².dx = dt -- (1)

Substituting the value of (2) in (1), we have

∫sin(x³).3x².dx = ∫ sin t.dt

Thus the integration of the above equation will give

∫ sin t.dt = −cos t + c

Again putting back the value of t from equation (ii), we get

∫ sin(x³).3x².dx = −cos(x³) + c

2.Integrate 2x.cos(x²−5) with respect to x.

Solution.

I = ∫2x cos (x²−5).dx

Let (x²–5) = t  -- (1)

Hence, 2x.dx = dt

Substituting these values, we have

I = ∫cos(t). dt

= sin t + C   --(2)

Substituting the value of (1) in (2), we have

I = sin (x²−5) + C

3. Find the integration of

Solution. Let t = tan^-1x -- (1)

 dt = (1/ 1+x² ) . dx

I = ∫ e^t . dt

= e^t + C --(2)

Substituting the value of (1) in (2), we have I = e^tan-1x + C. 

Q: What is the general form of integration by substitution?

Q: What should be kept in mind while choosing a term for substitution?

Q: Why is integration by substitution done?

Q: Explain the terms in the following equation: ∫ f(x) dx = F(x) + C

Q: What are the different methods of integration?