Integration by SubstitutionThe method of substitution can often be used to simplify complicated integration problems. It is based on the chain rule for differentiation.

Suppose that F is an antiderivative of f, and g is a differentiable function. Then

by the chain rule.

In integral form, we can write this as

( ( )) ( ( )) ( )d F g x F g x g xdx

′ ′=

( ( )) ( ) ( ( ))F g x g x dx F g x C′ ′ = +∫

( ( )) ( ) ( ( )) (1)f g x g x dx F g x C′ = +∫

or

To see how to use this we make a substitution, letting u = g(x). Then we see that Looking at formula (1) we see

So, formula (1) becomes

( ) .du g x dx′=

( ) ( ) (2)f u du F u C= +∫

( ( )) ( ) ( ( )) (1)f g x g x dx F g x C′ = +∫u du u

Definition: The process of turning (1) into (2) by making a substitution u = g(x), du = g′(x)dx is called the method of u-substitution.

Example. Evaluate the indefinite integral2 50( 1) (2 )x x dx+∫

Solution. We know how to handle powers of a variable, so the above formula suggests that we make the substitution

Then

2( 1)u x= + 2du xdx=

( )51251 12 50 50( 1) (2 )51 51

xux x dx u du C C

++ = = + = +∫ ∫

Here we had a fortunate situation where the division of the integral into du plus parts expressible in terms of u was immediate. Often some manipulation must take place.

Example. Evaluate the indefinite integral2 30( 1)x xdx+∫

Solution. Here we feel that we should make a similar substitution to the one made previously. We let

Then the integrand becomes u30, but we do not have duremaining. We need to express the rest of the integral in terms of u and du. Since we see that Then

2( 1)u x= + 2du xdx=

( )31231 11 1 12 30 30 30( 1)2 2 2 31 62

xux xdx u du u du C C

++ = = = + = +∫∫ ∫

2du xdx= 1 .2

xdx du=

In the method of u-substitution, you have control over the choice of u, but once that choice is made it must be possible to express all parts of the original integral, including the differential, in terms of u and du. There can be no extra x part remaining.

Extra terms involving x that might be left over cannot be brought outside the integral sign - they must be expressed in terms of u.

Example. Try to evaluate the indefinite integral

by the method of substitution.

3 30( 1)x xdx+∫

Solution. As before, we attempt the substitution

Then the integrand becomes u30, but there is no way to express the remaining xdx in terms of du. Some students would attempt to say that

3( 1)u x= + 23du x dx=

23 3 ( )du x dx x xdx= =

And therefore that

3duxdxx

=

Then the integral would become

3 30 30( 1)3dux xdx ux

+ =∫ ∫

This is hopeless since you cannot move the x term outside the integral sign.

In fact, the integral cannot be done at all by this method.

Example. What would work is to evaluate the indefinite integral

by the method of substitution.

3 30 2( 1)x x dx+∫

Solution. We attempt the substitution

This time the integrand becomes u30, and since

3( 1)u x= + 23du x dx=

Therefore the integral becomes:

123

x dx du=23du x dx= we have

31 3 311 1 ( 1)3 30 2 30 30( 1)3 3 93 93

u xx x dx u du u du C C++ = = = + = +∫ ∫ ∫

We summarize the steps in the method.

Integration by substitution.

Step1. Make a choice for u, say u = g(x)

Step2. Compute du/dx = g′(x)

Step3. Make the substitution u = g(x), du = g′(x)dx

At this point the entire integral must be in terms of u; no x’s should remain. If they do, try a different substitution.

Step4. Evaluate the resulting integral if possible.

Step5. Replace u by g(x), so that the final answer is in terms of x.

Example. Use the method of substitution to evaluate the indefinite integral

sin(5 )x dx∫

Solution. To reduce the function to a familiar one, we make the substitution 5u x= 5du dx=

Then we have so the integral becomes: 15

dx du=

1sin(5 ) sin( )

51 11 sin( ) cos( ) cos(5 )5 55

x dx u du

u du u C x C

=∫ ∫

= =− + =− +∫

Example. Use the method of substitution to evaluate the indefinite integral 2sin ( )cos( )x x dx∫

Solution. One must always ask the question - “If I choose this value for u, can I find du in the integral expression?”

We note that the differential of sin(x) is cos(x)dx, which is present in the above integral. This suggests that we make the substitution sin( )u x= cos( )du x dx=

Then the integral becomes:3 3sin ( )2 2sin ( )cos( )3 3

u xx x dx u du C C= = + = +∫ ∫

Example. Use the method of substitution to evaluate the indefinite integral 2sec (4 1)x dx+∫

Solution. A useful principle is- “Get rid of anything causing trouble by substituting for it”

In this integral, we would be fine if we had a variable in the sec2 term, rather than a more complicated expression. Thus we choose the substitution

4 1u x= + 4du dx=Then and the integral becomes:1

4dx du=

1 12 2 2sec (4 1) sec ( ) sec ( )4 4

1 1tan( ) tan(4 1)4 4

x dx u du u

u C x C

+ = =∫ ∫ ∫

= + = + +

Example. Use the method of substitution to evaluate the indefinite integral 4

2 5(2 7)( 7 3)x x x dx+ + +∫

Solution. Some quantity is being raised to a power. If we substitute for it, we can use the power rule for integration -provided that we can find its differential. We try the substitution:

2 7 3u x x= + + (2 7)du x dx= +The integral becomes:

4 4 9 95 52 5 5 25 5(2 7)( 7 3) ( 7 3)9 9

x x x dx u du u C x x C+ + + = = + = + + +∫ ∫

Example. Use the method of substitution to evaluate the indefinite integral

1

xedxxe

∫+

Solution. We might first consider the substitution .xdu e dx=

The integral becomes:

xu e=1

11

xedx dux ue

=∫ ∫ ++This integral could be evaluated by making a further substitution for u + 1. However, it is easier to just begin instead with the substitution: 1 xu e= + .xdu e dx=Then the integral becomes

Note. The absolute value could be omitted here, since the expression is always positive.1 xe+

( )1ln(| |) ln |1 )|

1

xe xdx du u C e Cx ue= = + = + +∫ ∫

+

Integration Hint

One thing to look for in an integrand that is a fraction is whether the numerator is equal to the derivative of the denominator (or is a multiple of it). In that case, substitute for the denominator.

The resulting integral will be of the form 1

duu∫

And this integrates to the natural logarithm of the absolute value of u.

Example. Use the method of substitution to evaluate the indefinite integral 2 1x xdx+∫

Solution. The problems in this integral seem to stem from the fact that the square root has a more complicated expression than x inside it. Thus we will get rid of this expression by substitution. Let u = 1 + x, du = dxThen the integral becomes:

( )5 3 1

2 2 2 2 2 21 ( 1) 2 1 2x xdx u udu u u udu u u u du+ = − = − + = − +∫ ∫ ∫ ∫

7 5 3 7 5 32 4 2 2 4 22 2 2 2 2 2(1 ) (1 ) (1 )7 5 3 7 5 3

u u u C x x x C= − + + = + − + + + +

5 3 1 7 5 322 22 2 2 2 2 22 257 3

u du u du u du u u u C = − + = − + +∫ ∫ ∫

Example. Use the method of substitution to evaluate the indefinite integral tan 2secxe xdx∫

Solution. The main problem in this integral is the presence of the expression tan x in the exponential. Since the differential of this expression is present we make the substitution.

Then the integral becomes:tanu x= 2secdu xdx=

tan 2 tansecx u u xe xdx e du e C e C= = + = +∫ ∫

Example. Use the method of substitution to evaluate the indefinite integral

24 5

xdx

x∫

−

Solution. The main problem in this integral appears to be the presence of the complicated expression under the square root. Let us try the substitution.

The expression xdx is then equal to so we have

24 5u x= − 10du xdx=−

1 11 1 1 12 2210 102 104 5

xdx du u du u

ux

− = − =− =−∫ ∫ ∫ −

110

du−

1 1 24 55 5

u C x C=− + =− − +

Example. Use the method of substitution to evaluate the indefinite integral 2cos(3 )x x dx∫

Solution. Let us substitute to get rid of the expression inside the cosine. Let

The expression xdx is then equal to so we have

23u x= 6du xdx=16

du

1 12cos(3 ) cos( ) cos( )6 6

x x dx u du u du= = ∫∫ ∫

1 1 2sin( ) sin(3 )6 6

u C x C= + = +

Note that in many ways, we were lucky in the previous two examples to have the differential of u present.

24 5

dx

x∫

−

2cos(3 )x dx∫

If those two integrals had been the apparently simpler integrals

and

then the first would require a more sophisticated substitution called a trigonometric substitution - we will learn this later-

and the second integration cannot be done at all, at least in terms of elementary functions (functions whose names you know from calculus I).

Similarly, we cannot express the integral 2xe dx−

∫

In terms of known functions. This integral is related by the fundamental theorem of calculus to the area under the curve

This curve plays a fundamental role in statistics. The indefinite integral can be evaluated numerically, and it is so important that it is given a name - the “error function” -and is written erf( )x

2xy e−=

Example. Use the method of substitution to evaluate the indefinite integral 2sec xdx

x∫

Solution. Let us substitute to get rid of the square root inside the trigonometric expression. Let

The expression is then equal to 2du, so we have

12u x x= =

11 22 2

dxdu x dxx

−= =

dxx

2sec 2 2sec ( )2 2 sec ( ) 2tan( ) 2tan( )xdx u du u du u C x Cx

= = = + = +∫ ∫ ∫