Standard Substitutions In Indefinite Integral Examples

To solve a differential equation or an integration problem, we have learned various types of approaches. Some fundamental integration rules are there in terms of solving cooperatively easier or basic integrals. However, solving integrals of a moderate level is not that easy for some of the students.

Let us start with a brief history of calculus. Lately, in the 17th century, calculus was invented. Some claim that Sir Isaac Newton was the inventor of calculus. While others credit German philosopher Gottfried Leibniz as the inventor of calculus. Apart from all the controversies, both the scientists have left their massive contributions to this field of mathematics and thus both the legends are given credit. The preliminary notion was that infinity can never be attained. The legends showed the world that nothing is impossible in the mathematical world. To be specific, attaining infinity is never possible. However, going closer and closer to it is desirable. Understanding the characteristics of certain functions at infinity is possible with the help of calculus or to be precise “limit”. Hence, calculus started its journey.

Now, coming to the point. Integrations are of two major types. Firstly definite integral and secondly indefinite integral. We can consider a definite integral as a subset of an indefinite integral. In the following, we will be learning about the substitution rules for indefinite integral.

Substitution rule for indefinite integral

We know that integration determines the area under a certain curve. Nonetheless, integral problems are solved easily by some simple tricks and techniques. One of the main methods of solving indefinite integral is the substitution method. To solve an easier integration problem, we may follow fundamental rules. But, if we try to solve a function that is itself a compound function (i.e., in this case, multiplication or division of two or more different functions) then we need to approach it differently. In that scenario, the substitution method is used in the integration. Let us try to understand how the rules work with a few examples.

   E.g._1: 

Let, f(x) = 2x cos(x²+1) and we have to find ∫ f(x)dx.

Here it is clearly seen that, f(x) is a product of 2 different functions. We suppose, g(x) = 2x and h(x) = cos(x²+1).

Now,

f(x) = g(x) × h(x)

Hence,

       ∫ f(x)dx = ∫g(x) × h(x) dx

Or, ∫ f(x)dx = ∫ 2x × {cos (x²+1)} dx

Now, let us assume that, x²+1 = z (where, z is another variable) … (eqn. 1)

Now, if we differentiate (eqn. 1) with respect to x, we get,

       d/dx (x²+1) = d/dx (z)

Or, 2x = dz/dx

Or, 2x dx = dz

(Always keep in mind that, functions are never multiplied with the integration factors, for example, “2x” is not multiplied with “dx”. However, we treat them multiplied in some cases according to our convenience.)

Then,

       ∫ f(x)dx = ∫ 2x × {cos (x²+1)} dx

Or, ∫ f(x)dx = ∫ 2x × cos (z) dx

Or, ∫ f(x)dx = ∫ cos (z) × (2x dx) [since, x²+1 = z]

Or, ∫ f(x)dx = ∫ cos (z) dz [since, dz = 2x dx]

Or, ∫ f(x)dx = sin (z) + c  [where, “c” is the integration constant]

Well, it may seem that the integration is completed here. Wait a minute. Have you been given the variable “z” in the problem? You only have been given “x” as the variable in the problem. Therefore, now you have to change the variable that you used in terms of substitution. Most of the students make a mistake here only. We have to put the value of “z” and end the solution.

Continuing,

Or, ∫ f(x)dx = sin (z) + c

Or, ∫ f(x)dx = sin (x²+1) + c

Therefore, solution to the given problem is, ∫ f(x)dx = sin (x²+1) + c [where, f(x) = 2x cos (x²+1)]

E.g._2

Let, f(x) = 6x²√x³+4 and we have to find ∫ f(x)dx.

Now, we have,

f(x) = 6x² × √x³+4

Let us apply the substitution, x³+4 = z

Then, by a similar approach, dz = 3x²dx (differentiating both sides with respect to x)

Hence,

       ∫ f(x)dx = ∫ 2 × √z × (3x²dx)

Or, ∫ f(x)dx = ∫ 2 × √z dz

Or, ∫ f(x)dx = 2 × ∫ √z dz

Or, ∫ f(x)dx = 4/3 × z√z +c (where c is the integration constant)

Or, ∫ f(x)dx = 4/3 × (x³+4) × √(x³+4) + c

Hence, the problem is solved.

With the above 2 examples, the significance of integration by substitution formula must be clarified. This method is also called the inverse method of chain rule of differentiation.

Have you noticed any similarity in the pattern of the functions that are solved by the method of substitution? Well, that similarity may amuse you. There must be a set pattern of functions where the substitution method is used in the integration. The integrable functions which are supposed to be solved by the substitution method must be of the form,

∫ f[g(x)]×[g’(x)]dx

General solution of the aforesaid integral can be found by the following steps.

Suppose,

       g(x) = y

Or, dy = g’(x)dx

Hence, ∫ f[g(x)]×[g’(x)]dx = ∫ f(y)dy and then solve the rest.

Conclusion

Have you read the entire article along with the given examples? Congratulations, you have learned the techniques of solving integration problems for a composite function. You have also learned a brief history of the invention of calculus. Always keep in mind that there will never be a formula to solve every problem in mathematics (or life, in general). Only your concept and wit will make you successful in solving any problem. Good luck and good day.