INTEGRATION 05

Integration by Substitution

Substitution simplifies an integral by replacing a complicated inner expression with a new variable.

The main idea

Substitution is the integration counterpart of the chain rule.

Suppose an integral contains

\[ f(g(x))g'(x). \]

Let

\[ u=g(x). \]

Then

\[ \boxed{ du=g'(x)\,dx } \]

and the integral becomes

\[ \int f(u)\,du. \]

1. Why substitution works

The chain rule says

\[ \frac{d}{dx}F(g(x)) = F'(g(x))g'(x). \]

\[ F'(u)=f(u), \]

then

\[ \frac{d}{dx}F(g(x)) = f(g(x))g'(x). \]

Reversing this relationship gives

\[ \boxed{ \int f(g(x))g'(x)\,dx = F(g(x))+C } \]

2. A simple example

Consider

\[ \int 2x(x^2+1)^3\,dx. \]

The inner expression is

\[ x^2+1. \]

Let

\[ u=x^2+1. \]

Differentiate:

\[ du=2x\,dx. \]

Therefore,

\[ \int 2x(x^2+1)^3\,dx = \int u^3\,du. \]

Integrate:

\[ \int u^3\,du = \frac{u^4}{4}+C. \]

Substitute back:

\[ \boxed{ \int 2x(x^2+1)^3\,dx = \frac{(x^2+1)^4}{4}+C } \]

3. The structure to look for

A useful pattern is

\[ \text{function of }g(x) \]

together with

\[ g'(x)\,dx. \]

For example,

\[ (3x+2)^5 \]

contains the inner function

\[ g(x)=3x+2. \]

Its derivative is

\[ g'(x)=3. \]

So a factor of \(3\) is exactly what substitution needs.

4. Example with a linear inner function

Evaluate

\[ \int (3x+2)^5\,dx. \]

Let

\[ u=3x+2. \]

Then

\[ du=3\,dx. \]

\[ dx=\frac{du}{3}. \]

Therefore,

\[ \int(3x+2)^5\,dx = \frac13\int u^5\,du. \]

Hence,

\[ = \frac13\cdot\frac{u^6}{6}+C. \]

\[ \boxed{ \int(3x+2)^5\,dx = \frac{(3x+2)^6}{18}+C } \]

5. Example with an exponential

Consider

\[ \int 2xe^{x^2}\,dx. \]

Let

\[ u=x^2. \]

Then

\[ du=2x\,dx. \]

So the integral becomes

\[ \int e^u\,du. \]

Therefore,

\[ \boxed{ \int 2xe^{x^2}\,dx = e^{x^2}+C } \]

6. Example with a logarithm

Consider

\[ \int \frac{2x}{x^2+1}\,dx. \]

Let

\[ u=x^2+1. \]

Then

\[ du=2x\,dx. \]

\[ \int \frac{2x}{x^2+1}\,dx = \int\frac1u\,du. \]

Therefore,

\[ \boxed{ \int \frac{2x}{x^2+1}\,dx = \ln(x^2+1)+C } \]

7. Example with trigonometric functions

Consider

\[ \int \cos(4x)\,dx. \]

Let

\[ u=4x. \]

Then

\[ du=4\,dx \]

\[ dx=\frac14\,du. \]

Therefore,

\[ \int\cos(4x)\,dx = \frac14\int\cos u\,du. \]

Thus,

\[ \boxed{ \int\cos(4x)\,dx = \frac14\sin(4x)+C } \]

8. Definite integrals and substitution

Substitution can also be used in definite integrals.

For example,

\[ \int_0^1 2x(x^2+1)^2\,dx. \]

Let

\[ u=x^2+1. \]

Then

\[ du=2x\,dx. \]

The limits must also change.

When

\[ x=0, \]

we have

\[ u=1. \]

When

\[ x=1, \]

we have

\[ u=2. \]

Therefore,

\[ \int_0^1 2x(x^2+1)^2\,dx = \int_1^2 u^2\,du. \]

\[ = \left[ \frac{u^3}{3} \right]_1^2 = \frac83-\frac13 = \frac73. \]

Hence,

\[ \boxed{ \int_0^1 2x(x^2+1)^2\,dx = \frac73 } \]

9. Two valid methods for definite integrals

After making a substitution in a definite integral, there are two valid approaches.

Either change the limits into the new variable \(u\),

or integrate in \(u\), substitute back to \(x\), and then use the original \(x\)-limits.

Do not mix \(u\)-expressions with \(x\)-limits.

10. Choosing the substitution

A good substitution usually replaces an inner expression whose derivative also appears in the integral.

\[ \boxed{ u=\text{inner expression} } \]

For example:

\[ (x^2+1)^5(2x) \]

suggests

\[ u=x^2+1. \]

And

\[ e^{3x} \]

suggests

\[ u=3x. \]

11. Not every integral fits immediately

Consider

\[ \int x(x^2+1)^3\,dx. \]

\[ u=x^2+1, \]

then

\[ du=2x\,dx. \]

The integral contains only \(x\,dx\), so write

\[ x\,dx=\frac12du. \]

Then

\[ \int x(x^2+1)^3\,dx = \frac12\int u^3\,du. \]

Thus,

\[ \boxed{ \int x(x^2+1)^3\,dx = \frac18(x^2+1)^4+C } \]

12. Why \(du\) is useful

\[ u=g(x), \]

then the differential is

\[ du=g'(x)\,dx. \]

This allows the original integral to be rewritten entirely in terms of \(u\).

\[ \boxed{ u=g(x) \qquad\Rightarrow\qquad du=g'(x)\,dx } \]

The differential notation makes the chain-rule structure especially visible.

13. The connection with the chain rule

Differentiation:

\[ F(g(x)) \xrightarrow{\ d/dx\ } F'(g(x))g'(x) \]

Integration by substitution reverses this:

\[ F'(g(x))g'(x) \xrightarrow{\ \int\ } F(g(x))+C. \]

14. Practical method

\[ \boxed{ 1.\; \text{Choose }u=g(x) } \] \[ \boxed{ 2.\; \text{Find }du=g'(x)\,dx } \] \[ \boxed{ 3.\; \text{Rewrite the entire integral in }u } \] \[ \boxed{ 4.\; \text{Integrate} } \] \[ \boxed{ 5.\; \text{Substitute back, or use transformed limits} } \]

The goal is to replace a complicated integral with a simpler one whose antiderivative is already known.