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Reverse Chain Rule

  • Last Updated : 14 Jul, 2021

Integrals are an important part of the theory of calculus. They are very useful in calculating the areas and volumes for arbitrarily complex functions, which otherwise are very hard to compute and are often bad approximations of the area or the volume enclosed by the function. Integrals are the reverse of the differentiation and that’s why they are called anti-derivatives. There are formulas for the integrals of standard functions, but these formulas usually do not extend to the cases where functions become complex. One such rule is the substitution rule. Let’s study this rule in detail. 

Reverse Chain Rule

The chain rule for derivatives allows to calculate the derivatives for very complex functions that involve one or more standard basic functions. For integration in such scenarios, there are multiple tricks or methods to simplify the calculations. For certain special classes of functions, the reverse chain rule is used. This rule is also called the “substitution rule” or the “u-substitution rule”. To use this rule, the function should have the form given below, 

∫f(g(x))g'(x)dx

Notice that in the function of the given form, both the function g(x) and its derivative g'(x) are present. Some examples of such functions are 

∫cos(x2).2xdx



In this case, g(x) = x2, f(x) = cos(x) and g'(x) = 2x. Now to calculate integrals for such functions using the substitution rule, a little modification is required. Consider the general form of such functions: 

∫f(g(x))g'(x)dx

Let u = g(x), differentiate this w.r.t x, 

u = g(x) \\ = \frac{du}{dx} = g'(x) \\ = du = g'(x)dx

Using this result in the previous equation, 

∫f(u)du

Now this integral can be calculated and finished by substituting the value of u in the final result. 

∫f(g(x))g'(x)dx = ∫f(u)du, where u = g(x) 



Calculating integrals using the Reverse Chain Rule

It is essential to look for the possible functions and their derivatives in the given integral functions. The natural question that comes to mind is how to know about the correct substitution. There are no rules for determining the correct substitution. Intuition for recognizing the correct substitution gets developed with practice. 

Example 1: Calculate the integral for ∫cos(x2).2xdx. 

Answer: 

F(x) = ∫cos(x2).2xdx

Look for a function, and it’s derivative in the whole function. In this case, 

g(x) = x2 and g'(x) = 2x 

Substituting, u = g(x) 

⇒ u = x2

⇒ du = 2xdx 

F(x) = ∫cos(u)du



⇒F(x) = sin(u) + C

⇒F(x) = sin(x2) + C

Now consider a different type of integral: 

Example 2: Calculate the integral for ∫tan(x)dx. 

Answer: 

F(x) = ∫tan(x)dx

This can be re-written as, 

F(x) = ∫\frac{sin(x)}{cos(x)}dx

Notice, here g(x) = cos(x) and g'(x) = -sin(x)

Substituting, u = g(x) 

⇒ u = cos(x)

⇒ du = -sin(x)dx 

⇒ -du = sin(x)dx

F(x) = ∫\frac{sin(x)}{cos(x)}dx

⇒ F(x) = ∫\frac{-1}{u}du

⇒ F(x) = -ln(u) + C 

⇒ F(x) = -ln(cos(x)) + C

Mistakes to avoid while using the Reverse-Chain Rule

One should be careful while choosing the “u” function from the original integral function. If it is not chosen carefully, the integrals may become overly complex rather than simplifying themselves. Also, in some scenarios, the u-functions are not immediately visible. The following list presents some things which should be kept in mind while solving the integral problems with the reverse-chain rule. 

1. For applying the reverse chain rule, the integral must be re-written in the form, 

w(u(x)).u'(x)



Where the u-function is the inner function of the composite factor. 

2. While integrating the composite function, the outer function should only be integrated after properly substituting the u-function and its derivatives. 

3. Sometimes, in some situations, the integral must be divided/multiplied by some constant factors of variables or the function may need rearranging. For example, 

F = ∫tan(x)dx

In this case, the u-function is not completely clear. So, tan(x) is rewritten in terms of sin(x) and cos(x). 

F = \int \frac{sin(x)}{cos(x)}dx

Now it is clear that cos(x) is the u-function. 

Sample Problems

Question 1: Calculate the integral for ∫cos(ex).exdx. 

Answer: 

F(x) = ∫cos(ex).exdx

Look for a function, and it’s derivative in the whole function. In this case, 

g(x) = ex and g'(x) = ex 

Substituting, u = g(x) 

⇒ u = ex

⇒ du = exdx 

F(x) = ∫cos(u)du

⇒F(x) = sin(u) + C

⇒F(x) = sin(ex) + C

Question 2: Calculate the integral for ∫(6x + 4)6.6dx. 

Answer: 



F(x) = ∫(6x + 4)6.6dx.

Look for a function, and it’s derivative in the whole function. In this case, 

g(x) = 6x + 4 and g'(x) = 6 

Substituting, u = g(x) 

⇒ u = 6x + 4

⇒ du = 6dx 

F(x) = ∫u6du

⇒F(x) = \frac{u^7}{7} + C

⇒F(x) = \frac{(6x + 4)^7}{7}+ C

Question 3: Calculate the integral for ∫(x2+ 1)3.2xdx. 

Answer: 

F(x) = ∫(x2 + 1)3.2xdx.

Look for a function, and it’s derivative of that function in the original function. In this case, 

g(x) = x2 + 1 and g'(x) = 2x 

Substituting, u = g(x) 

⇒ u = x2 + 1

⇒ du = 2xdx 

F(x) = ∫u3du

⇒F(x) = 4\frac{u^4}{4} + C

⇒F(x) = \frac{(x^2 + 1)^4}{4}+ C



Question 4: Calculate the integral for \int \frac{x}{x^2 +1}dx

Answer: 

F(x) = \int \frac{x}{x^2 +1}dx

Look for a function, and it’s a derivative of that function in the original function. In this case, 

g(x) = x2 + 1 and g'(x) = 2x 

Substituting, u = g(x) 

⇒ u = x2 + 1

⇒ du = 2xdx 

F(x) = \int \frac{1}{2u}du

⇒F(x) = \frac{ln(u)}{2} + C

⇒F(x) = \frac{ln(x^2 + 1)}{2} + C

Question 5: Calculate the integral for \int \frac{1}{x +3}dx

Answer: 

F(x) = \int \frac{1}{x +3}dx

Look for a function, and it’s a derivative of that function in the original function. In this case, 

g(x) = x + 3 and g'(x) = 1 

Substituting, u = g(x) 

⇒ u = x + 3

⇒ du = dx 

F(x) = \int \frac{1}{u}du



⇒F(x) = ln(u) + C

⇒F(x) = ln(x + 3) + C

Question 6: Calculate the integral for \int \frac{(ln(x))^2}{x}dx

Answer: 

F(x) = \int \frac{(ln(x))^2}{x}dx

Look for a function, and it’s derivative in the whole function. In this case, 

g(x) = ln(x) and g'(x) = \frac{1}{x}

Substituting, u = g(x)

⇒ u = ln(x)dx

⇒ du = \frac{1}{x}dx      

F(x) = ∫u2du

⇒F(x) = \frac{u^3}{3} + C

⇒F(x) = \frac{(ln(x))^3}{3} + C

Question 7: Calculate the integral for ∫sin(1-x)cos(1-x)dx

Answer: 

F(x) = ∫sin(1-x)cos(1-x)dx

This function can be re-written as, 

F(x) = \frac{sin2(x-1)}{2}dx

Look for a function, and it’s derivative in the whole function. In this case, 

g(x) = 2(x -1) and g'(x) = 2



Substituting, u = g(x)

⇒ u = 2(x – 1)

⇒ du = 2dx 

F(x) = \frac{sin(u)}{4}du

⇒F(x) = \frac{-cos(u)}{4} + C

⇒F(x) = \frac{-cos(2(x - 1))}{4} + C

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