Integration by Substitution in Calculus with Concept and Steps

In order to determine the integrals of function accurately, we are required to develop techniques that can minimize the functions to standard form. The integration by substitution class 12th is one important topic which we will discuss in this article.

In the integration by substitution,a given integer f (x) dx can be changed into another form by changing the independent variable x to z. 

This is done by substituting x = k(z).

Consider I = f(x)dx

Now substitute x = k(z) so that dx/dz = k’(z) or dx = k’(z) dz.

Hence, I = \[\int\] f(x) dx = f[k(z) k’(z)dz

It is essential to notice here that you should make a substitution for a function whose derivative also appears in the integrals as shown in the below -solved examples.

What is Integration?

The integration represents the summation of discrete data. The integral is usually calculated to find the functions which detail information about the area, displacement, volume, which appears due to the collection of small data, which cannot be measured singularly. Generally, in calculus, the idea of limit is used where algebra and geometry are applied. Limits assist us in the study of the result of points on a graph such as how they get nearer to each other until their distance is almost zero. There are two major types of calculus –

• Differential Calculus

• Integral Calculus

Integration by Substitution Method

In the integration by substitution method, any given integral can be changed into a simple form of integral by substituting the independent variable by others.

For example,

Let us consider an equation having an independent variable in z, i.e. 

\[\int\] sin (z³).3z².dz———————–(i),

The independent variable given in the above example can be changed into another variable say k.

Substituting z³ = k ———————-(ii)

By differentiation of the above equation, we get

3z².dz = dk———————-(iii)

Substituting the value of equation (ii) and (iii) in equation (i), we get

\[\int\] sin (z³).3z².dz = \[\int\] sin k.dk

Hence, the integration of the above equation will give us

\[\int\] sin k.dk= - cos k + c

Again substituting back the value of k from equation (ii), we get

\[\int\]sin (z³).3z².dz = -cos(z³ )+ c

The standard form of integration by substitution is:

\[\int\]f(g(z)).g'(z).dz = f(k).dk, where k = g(z)

The integration by substitution method is extremely useful when we make a substitution for a function whose derivative is also included in the integer. With this, the function simplifies and then the basic integration formula can be used to integrate the function.

Integration Examples

1. Integrate sin(zx)  in terms to x,

Solution: We know that the derivative of zx = z

No, let us substitute zx = k son than zdx = dk

Hence,

\[\int\]Sin zx dx = 1/z sink dk

= -1/z cos k + C

= -1/z cos zx + C

2. Integrate 2x sin (x² + 1) in terms of x.

Solution: As, we know that the derivative of (x² +1) = 2x

Now, let us substitute x + 1= k so that 2x dx = dk. Hence,

\[\int\]2x sin (x²+1) dx = \[\int\]sin k dk

= - Cos k + C

= - Cos (x² +1) + c

Solved Examples

1. Integrate 2x cos (x² - 5) in terms to x.

Solution:

I = \[\int\]2xcos (x² - 5).dx

Let  x² - 5 = t…………….(1)

2x.dx = dt

Substituting these values, we have

I = \[\int\]cos(t).dt

= sin t + C……………...(2)

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

= sin (x² - 5) + C.

2. \[\int\] X / \[\sqrt{X + 1}\] dx

Solution: Let u = x + 1

Hence, x= u-1

dx/du = 1 so dx = du

Hence, the integral becomes

\[\int\] u-1/\[\sqrt{u}\]du

= \[\int\] u/\[\sqrt{u}\]-1/\[\sqrt{u}\]

= u1/2

\[\int\]u1/2 – u1/2 du

= 2u3/2 - 2 u1/2  + C

We will now substitute the values of x’s back in

2 (x + 1)3/2 -2(x+1)1/2  + C

Quiz Time

1. What should be used for u in the integral?

1. No u required

2. 4t²

3. T³

4. 1/4t²

2.  What should be assigned to u in the integral?

1. 2x²

2. sin(2x²)

3. 5x

4. No u required

3. \[\int\] 6x(x² +1)2 dx

1. (x² +1)³ + C

2. 3 (x² +1)³ + C

3. (x² +1)1 + C

4. 6(x² +1)³ + C