Subject: Calculus

# Substitution Rule

Substitution rule is a method of solving a complicated integral or antiderivatives. This works like a charm especially for integral functions that are composite. Substitution rule is very much alike to inverse chain rule method. However, inverse chain rule method has problems.

## The Problem

If you have an integral of the form

then how would you solve that using inverse chain rule method and other previous methods? We still can't solve that one with current knowledge of doing integration that's why let us study substitution rule to help us solve such integral.

## The solution and how it works...

Let us find solution to the integral depicted above by substitution technique. Let u=g(x), then let us differentiate u by,

Putting this into the integral above, we get

See! By doing substitution method, we simplify the integral consisting of a complicated function because of the composite function f(g(x)). The integral now looks very simple, clean and easy to integrate.

### Illustration

As with all of Mathematics and Sciences, the best way to start to get your head around a new concept is to look and study many illustrations and examples and the most effective way to be able to apply the new idea is to do many problems and exercises. Let before doing your own exercise, let me give you an illustration on how to do substitution method.

Say, we are to evaluate the following integral

You can suggest any method but I will do substitution method. Let u=2+x^2 such that du=2xdx. This gives xdx=\frac{du}{2}.

Thus, our integral would become

Putting back u=2+x^2, we get

### Example #1

Evaluate: \int e^{\cos{x}}\sin{x}dx

We let \cos{x}=u then du=-sin{x}dx, such that the integral above would take the form

And we know that

Putting back \cos{x}=u, then

### Example #2

Calculate the following integral using substitution rule: \int 2x\sqrt{x^2+5}dx

ANSWER: \int 2x\sqrt{x^2+5}dx=\frac{2}{3}(x^2+5)^{3/2}+C

### Example #3

Solve using substitution method: \int \tan{x}dx

We know that

\tan{x}=\frac{\sin{x}}{\cos{x}}

Thus,

\int \tan{x}dx=\frac{\sin{x}}{\cos{x}}dx

Let u=\cos{x} such that du=-\sin{x}dx. Therefore,

\int \tan{x}dx=-\int \frac{1}{u}du

which is very easy to solve.

-\int \frac{1}{u}du=-\ln{u} +C

Putting back u= \cos{x}, we get

\ln{\frac{1}{u}}+C=-\ln{\cos{x}}+C

Thus,

\int \tan{x}dx=-\ln{\cos{x}}+C

### Example #4

Solve: \int (1-\frac{1}{s})e^{s-\ln{s}}ds

ANSWER: \int (1-\frac{1}{s})e^{s-\ln{s}}ds=e^{s-\ln{s}}+C

### Example #5

Integrate: \int -\sin{w}\ln{|\cos{w}|}dw

ANSWER: \int -\sin{w}\ln{|\cos{w}|}dw=\cos{w}\ln{|\cos{w}|}-\cos{w}+C

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