U-Substitution - UC Davis Math

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U-Substitution

THE METHOD OF U-SUBSTITUTION

The following problems involve the method of u-substitution. It is a method for finding antiderivatives. We will assume knowledge of the following well-known, basic indefinite integral formulas :

$\displaystyle{ \int x^n \,dx } = \displaystyle{ {x^{n+1} \over n+1 } + C }$
$\displaystyle{ \int { 1 \over x } \,dx } = \displaystyle{ \ln \vert x\vert + C }$
$\displaystyle{ \int e^x \,dx } = \displaystyle{ e^x + C }$
$\displaystyle{ \int a^x \,dx } = \displaystyle{ { a^x \over \ln a } + C }$ , where a is a constant
$\displaystyle{ \int \cos x \,dx } = \displaystyle{ \sin x + C }$
$\displaystyle{ \int \sin x \,dx } = \displaystyle{ -\cos x + C }$
$\displaystyle{ \int k f(x) \,dx } = k \displaystyle{ \int f(x) \,dx }$ , where k is a constant
$\displaystyle{ \int ( f(x) \pm g(x) ) \,dx } = \displaystyle{ \int f(x) \,dx } \pm \displaystyle{ \int g(x) \,dx }$

The method of u-substitution is a method for algebraically simplifying the form of a function so that its antiderivative can be easily recognized. This method is intimately related to the chain rule for differentiation. For example, since the derivative of ex is

$D \{ e^x \} = e^x$ ,

it follows easily that

$\displaystyle{ \int e^x \,dx } = e^x + C$ .

However, it may not be obvious to some how to integrate

$\displaystyle{ \int (2x+2) e^{ x^2 + 2x + 3 } \,dx }$ .

Note that the derivative of $\displaystyle{ e^{ x^2 + 2x + 3 } }$ can be computed using the chain rule and is

$\displaystyle{ D \{ e^{ x^2 + 2x + 3 } \} = e^{ x^2+2x+3 } \ D \{ x^2+2x+3 \} }$

$= \displaystyle{ e^{ x^2+2x+3 } (2x+2) }$

$= \displaystyle{ (2x+2) e^{ x^2+2x+3 } }$ .

Thus, it follows easily that

$\displaystyle{ \int (2x+2) e^{ x^2+2x+3 } \,dx } = e^{ x^2+2x+3 } + C$ .

This is an illustration of the chain rule "backwards". Now the method of u-substitution will be illustrated on this same example. Begin with

$\displaystyle{ \int (2x+2) e^{ x^2 + 2x + 3 } \,dx }$ ,

and let

u = x2+2x+3 .

Then the derivative of u is

$\displaystyle{ { du \over dx } } = 2x+2$ .

Now "pretend" that the differentiation notation $\displaystyle{ { du \over dx } }$ is an arithmetic fraction, and multiply both sides of the previous equation by dx getting

$\displaystyle{ { du \over dx } \ dx } = (2x+2) dx$

du = (2x+2) dx .

Make substitutions into the original problem, removing all forms of x , resulting in

$\displaystyle{ \int (2x+2) e^{ x^2+2x+3 } \,dx } = \displaystyle{ \int e^{ x^2+2x+3 } (2x+2) \,dx }$

$= \displaystyle{ \int e^{ u } \,du }$

= e u + C

= e x2+2x+3 + C .

Of course, it is the same answer that we got before, using the chain rule "backwards". In essence, the method of u-substitution is a way to recognize the antiderivative of a chain rule derivative. Here is another illustraion of u-substitution. Consider

$\displaystyle{ \int { x^2+1 \over x^3+3x } \,dx }$ .

Let

u = x3+3x .

Then (Go directly to the du part.)

du = (3x2+3) dx = 3(x2+1) dx ,

so that

(1/3) du = (x2+1) dx .

Make substitutions into the original problem, removing all forms of x , resulting in

$\displaystyle{ \int { x^2+1 \over x^3+3x } \,dx } = \displaystyle{ \int { 1 \over x^3+3x } \ (x^2+1) \,dx }$

$= \displaystyle{ \int { 1 \over u } (1/3) \,du }$

$= \displaystyle{ (1/3) \int { 1 \over u } \,du }$

$= (1/3) \ln\vert u\vert + C$

$= (1/3) \ln\vert x^3+3x\vert + C$ .

Most of the following problems are average. A few are challenging. Make careful and precise use of the differential notation dx and du and be careful when arithmetically and algebraically simplifying expressions.

PROBLEM 1 : Integrate $\displaystyle{ \int { (2x+5) (x^2+5x)^7 } \,dx }$ .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Integrate $\displaystyle{ \int { (3-x)^{10} } \,dx }$ .
Click HERE to see a detailed solution to problem 2.
PROBLEM 3 : Integrate $\displaystyle{ \int \sqrt{ 7x+9 } \,dx }$ .
Click HERE to see a detailed solution to problem 3.
PROBLEM 4 : Integrate $\displaystyle{ \int { x^3 \over (1+x^4)^{1/3} } \,dx }$ .
Click HERE to see a detailed solution to problem 4.
PROBLEM 5 : Integrate $\displaystyle{ \int { e^{5x+2} } \,dx }$ .
Click HERE to see a detailed solution to problem 5.
PROBLEM 6 : Integrate $\displaystyle{ \int { 4 \cos(3x) } \,dx }$ .
Click HERE to see a detailed solution to problem 6.
PROBLEM 7 : Integrate $\displaystyle{ \int { \sin( \ln x ) \over x } \,dx }$ .
Click HERE to see a detailed solution to problem 7.
PROBLEM 8 : Integrate $\displaystyle{ \int { 3x+6 \over x^2+4x-3 } \,dx }$ .
Click HERE to see a detailed solution to problem 8.
PROBLEM 9 : Integrate $\displaystyle{ \int { x \ 3^{ x^2+1} } \,dx }$ .
Click HERE to see a detailed solution to problem 9.
PROBLEM 10 : Integrate $\displaystyle{ \int { 3 \over x \ln x } \,dx }$ .
Click HERE to see a detailed solution to problem 10.
PROBLEM 11 : Integrate $\displaystyle{ \int { \cos(5x) \over e^{ \sin(5x) } } \,dx }$ .
Click HERE to see a detailed solution to problem 11.
PROBLEM 12 : Integrate $\displaystyle{ \int_{0}^{ \sqrt{\pi} } { x \sin(x^2) } \,dx }$ .
Click HERE to see a detailed solution to problem 12.

The following problems require u-substitution with a variation. I call this variation a "back substitution". For example, if u = x+1 , then x=u-1 is what I refer to as a "back substitution".

PROBLEM 13 : Integrate $\displaystyle{ \int { (x+3) (x-1)^5 } \,dx }$ .
Click HERE to see a detailed solution to problem 13.
PROBLEM 14 : Integrate $\displaystyle{ \int { x \sqrt{4-x} } \,dx }$ .
Click HERE to see a detailed solution to problem 14.
PROBLEM 15 : $\displaystyle{ \int { x+5 \over 2x+3 } \,dx }$ .
Click HERE to see a detailed solution to problem 15.
PROBLEM 16 : $\displaystyle{ \int { x^2+4 \over x+2 } \,dx }$ .
Click HERE to see a detailed solution to problem 16.
PROBLEM 17 : $\displaystyle{ \int { \cos(5^x) \ln( \sin(5^x) ) \over \sin(5^x) } \,dx }$ .
Click HERE to see a detailed solution to problem 17.
PROBLEM 18 : Integrate $\displaystyle{ \int_{0}^{4} { x \sin(x^2) } \,dx }$ .
Click HERE to see a detailed solution to problem 18.