Over the past five chapters we have seen that the process of finding

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The Substitution Rule

O

ver the past five chapters we have seen that the process of finding indefinite integrals (that is, the process of integration) is essential in calculus. For this, we have so far relied on a relatively sparse list of integration rules that come from reversing differentiation rules.

Here is a summary of our integration rules so far. The only difference between this list and the one on page 181 is that here the variablex has been replaced byu. (For a reason that will soon become apparent.)

Integration Rules R c du = cu+C R undu = _n1 +1un+1+C Z₁ udu = ln|u| +C R eudu = eu+C R budu = _ln(b)1 bu+C R sin(u) du = °cos(u)+C R cos(u) du = sin(u)+C R_sec₂_{(u) du} = tan(u)+C R csc2(u) du = °cot(u)+C R sec(u)tan(u) du = sec(u)+C R csc(u)cot(u) du = °csc(u)+C Z ₁ p 1°u2 du = sin°1(u)+C Z ₁ 1+u2 du = tan°1(u)+C Z ₁ upu2_°₁du = sec°1|u| +C

Recall that each of these integration rules is verified by differentiating the answer and checking that the derivative is the same as the integrand.

Z

f (u) du = F(u)+C d

du

Keep the above list of integration rules in mind, for we will refer to it many times in this chapter.

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The above list does not contain the reverse of every differentiation rule. For instance, the reverse of the quotient rule d

du

h_{f (u)}

g(u)

i

= f0(u)g(u)°f (u)g° 0(u)

g(u)¢2 is

Z_f0_(u)g(u)_°_{f (u)g}0_(u)

°

g(u)¢2 du = f (u) g(u)+C.

This “rule” is useless (and is not on our list) because we rarely if ever need to find the integral of a function that has the precise form f0_{(u)g(u)°f (u)g}0_(u)

°

g(u)¢2 .

But there is one differentiation rule whose reverse is extremely useful. That rule is the chain rule. The reverse of the chain rule is an integration rule called the substitution rule. That is this chapter’s topic.

45.1 The Substitution Rule

ProvidedF and gare differentiable, the chain rule (for differentiation) says d

dx h

F°g(x)¢i = F0°_g(x)¢_·_g0_(x).

IfF0_(x)₌_{f (x)}_{, that is, if}_F _{is an antiderivative of} _f_{, then this becomes}

d dx

h

F°g(x)¢i = f°g(x)¢·g0_(x).

In other words, ifF is an antiderivative of f, then Z

f°g(x)¢·g0(x) dx = F°g(x)¢+C. (§) This is almost our new rule, but first we are going to simplify it. Let

g(x)=u, so g0_(x)₌ du

dx. Multipling both sides of this by the differentialdx,

g0_{(x) dx}₌_du_{. Using these boxed equations, replace (or substitute) the} _g(x)

in Equation (§) withu, and the g0_{(x) dx}_with _du_:

Z

f°g(x)¢g0_{(x) dx}₌_F°_g(x)¢₊_C

Z

f (u) du=F(u)+C

Becauseg(x)=u, the right-hand sides of these equations are equal. Thus left-hand sides are equal too. That is our new rule.

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The substitution rule can convert a complicated integral to a simple one.

Substitution Rule If u=g(x), then Zf°g(x)¢g0_{(x) dx} ₌ Z_{f (u) du}_.

| {z }

complicated |simple{z } When faced with a complicated integral to which no integration rules apply, the substitution rule—when applicable—lets us make a substitution u= g(x) that converts the integral to a simpler formRf (u) du to which a rule may apply. This is somewhat of an art, and occasionally involves some guesswork and trial and error. (And it definitely requires some practice and experience!) Following are some examples of the substitution rule.

Example 45.1 FindZsin°x2¢2x dx.

Solution This is does not match any of our familiar integration rules, but

the closest match is the ruleRsin(u) du= °cos(u)+C. To make the present problem more like this rule, we can make the substitutionu=x2. With this,

Z

sin°x2¢2x dx = Z

sin(u) 2x dx

But we still have the 2x dx where we want just du. (Remember, we are aiming forRsin(u) du.) But also we choseu=x2, and this means du_dx =2x. From this we get du=2x dx. Thus the above2x dxactually equals du, and the above turns into the simple integral

= Z

sin(u) du = °cos(u)+C.

But sinceu=x2, theuin°cos(u)+Cis actually x2. We have our answer.

Answer: Zsin°x2¢2x dx = °cos°x2¢+C.

To check this answer, notice that _dxd h°cos°x2¢+Ci = °sin°x2¢2x.

This example illustrates a general approach to using the substitution rule to find an indefinite integral that does not match any known rule: Think of an integration rule that most closely matches the problem. Then make a substitution u= g(x) that moves the problem closer to that rule. Once this choice is made, then du/dx= g0_(x)_{, so} _du₌ _g0_(x)dx_{. Look for a}

g0_(x)dx _{in the problem and replace it with}_du_{. If you now have something}

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Example 45.2 FindZsec

°

ln(x)¢tan°ln(x)¢

x dx.

Solution This integral is similar to the ruleRsec(u)tan(u) du=sec(x)+C. To make it match this rule, use u=ln(x) so du_dx= 1_x, hence du= 1_xdx . Then

Z_sec°_ln(x)¢_tan°_ln(x)¢ x dx = Z sec°ln(x)¢tan°ln(x)¢1 xdx = Z sec(u)tan(u) du = sec(u)+C = sec°ln(x)¢+C

We can check the the answer we just got by differentiating it to see if we get the integrand. By the chain rule,

d dx

h

sec°ln(x)¢+Ci = sec°ln(x)¢tan°ln(x)¢1 x+0 =

sec°ln(x)¢tan°ln(x)¢

x .

This is indeed the integrand, so our answer is correct. You can check all the examples in this section the same way.

Example 45.3 FindZ₂eppx_x dx.

Solution This integral is similar to the ruleReudu=eu+C. To make it match this rule, use u=px so du_dx=_2px1 , hence du= _2px1 dx . Then

Z _epx 2px dx = Z epx 1 2px dx = Z eudu = eu+C = epx+C. Example 45.4 FindZpex_ex_dx_.

Solution This isZ°ex¢1/2exdx, which resemblesZu1/2du, to which the power rule applies. So let u=ex _{. Then} du

dx=ex, hence du=exdx . Then Zp ex_ex_dx ₌ Z°_ex¢1/2_ex_dx = Z u1/2du = _1/21 +1u 1/2+1₊_C ₌ 2 3 p u3+C = 2₃pex3₊_C.

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Example 45.5 FindZ_x₃3x2+8x°1

+4x2°x+1 dx.

Solution Notice that the numerator of the integrand happens to be

the derivative of the denominator. Thus suggests u=x3+4x2°x+1 , so

du dx=3x2+8x°1, and hence du= ° 3x2+8x°1¢dx . Then Z _3x2₊_8x_°₁ x3₊_4x2_°_x₊₁dx = Z ₁ x3₊_4x2_°_x₊₁ ° 3x2+8x°1¢ dx = Z₁ udu = ln|u| +C = lnØØ_x3₊_4x2_°_x₊₁Ø_{Ø +}_C.

The substitution rule applies only to integrals that have the exact form R

f°g(x)¢·g0_{(x) dx}_{, or those that can be put into this form algebraically. Once}

the substitutionu=g(x)is made, the integral has the simpler formRf (u) du. After some practice, when confronted with an integral to which substitution applies you may immediately see the answer without actually doing the substitution. That is good.

But often the structureRf°g(x)¢·g0_{(x) dx}_{is not immediately apparent,}

and you may have to do some intelligent guessing to come up with workable substitutionu=g(x). Sometimes further algebraic manipulation is needed to attain a familiar formRf (u) du. We now work some examples of this type.

Example 45.6 FindZsin°x2¢x dx.

Solution This looks almost identical to Example 45.1. And as in that

example, the ruleRsin(u) du= °cos(u)+Cseems to be the closest match, so try u=x2 . From this du_dx=2x, so du=2x dx . But here we have a problem. We’d like to replace thex dxat the end ofRsin°x2¢_{x dx}_with_du_{, but}_du₆₌_{x dx}_.

The du=2x dx that came from our choiceu=x2 doesn’t match the x dxin the integral. To fix the problem, just dividedu=2x dxby 2 to get 1₂du=x dx. Now we know thex dxactually equals 1₂du. Making these substitutions,

Z sin°x2¢x dx = Z sin(u)1 2du = 1₂ Z

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Example 45.7 FindZsec2°px

¢_tan₂°p_x¢

p_x dx.

Solution After some thought, you may hit upon this idea: u=tan°px¢ . By the chain rule du

dx =sec2(px)2px1 =sec

2(px)

2px . Solve this as 2 du=sec_px2(px)dx . Now we have Z_sec2°p_x¢_tan2°p_x¢ p_x dx = Z°tan°px¢¢2sec2 °p x¢ p_x dx = Z u22 du = 2Zu2du = 2u3 3 +C = 2 ° tan°px¢¢3 3 +C = 2 tan3°px¢ 3 +C

Example 45.8 FindZcos(x) sin(x) dx.

Solution Choose u=cos(x) , sodu

dx = °sin(x). Solve this as °du=sin(x) dx

to get a match for thesin(x) dxthat appears in the integral. Then Z cos(x) sin(x) dx = Z u(°du) = ° Z u du = °u 2 2 +C = ° cos2_(x) 2 +C. That is our answer, but here is another approach. Choose u=sin(x) , so

du

dx=cos(x). Solve this as du=cos(x) dx . With these substitutions we get

Z cos(x) sin(x) dx = Z sin(x) cos(x) dx = Z u du = u 2 2 +C = sin2(x) 2 +C. So we got an answer in two different ways. But the answers look different. To see why this is not a mistake, notesin2(x)+cos2(x)=1, sosin2(x)=1°cos2(x). With this, our second answer is

sin2_(x) 2 +C = 1°cos2(x) 2 +C = ° cos2_(x) 2 + 1 2+C.

So the two answer are different: one is just the other plus 1/2. Since we regard the constant as arbitrary, we just write it asCin both cases.

Next we’ll do two examples that are very simple; so simple that students often make hasty mistakes when doing integrals of this type. We will use substitution to solve them here, but after some practice you will do this kind of problem mentally, getting the answer in one step.

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Example 45.9 FindZe°x_dx_.

Solution This closely resembles the ruleReudu=eu+C, so to convert to that form we choose u= °x . Then du_dx= °1, so dx= °du . Making these substitutions,Ze°x_dx₌Z_eu₍_°_du)_{= °}Z_eu_du_{= °}_eu₊_C_{= °}_e°x₊_C.

Example 45.10 FindZcos(2x) dx.

Solution This closely resembles the rule Rcos(u) du=sin(u)+C, so to convert to that form we choose u=2x . Then du_dx=2, so dx=1₂du . Then Z cos(2x) dx= Z cos(u)1 2du= 1 2 Z cos(u) du=1₂sin(u)+C. Example 45.11 FindZxpx+1 dx.

Solution Since this looks somewhat likeRpu du (which we can do with the power rule for integration), put u=x+1 . Then du_dx =1, so du=dx .

Then Z xpx+1 dx = Z x(x+1)1/2dx = Z x u1/2du.

This looks fine, except that there is still an xin the picture, and it didn’t go away with the substitution fordu. To take care of this, notice because we have made the substitutionu=x+1, it follows thatx=u°1. With this our computation continues as Z x u1/2du = Z (u°1) u1/2du = Z≥(u·u1/2°u1/2¥du = Z≥u3/2°u1/2¥du = _3/21 +1u 3/2+1₊ 1 1/2+1u 1/2+1₊_C = 2₅u5/2+2₃u3/2+C = 2₅pu5+2₃pu3+C = 2₅px+15+2₃px+13+C

It is important to practice some exercises before moving to the next section.

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45.2 Substitution in Definite Integrals

Part 2 of the fundamental theorem of calculus states Z_b

a f (x) dx =

h F(x)ib

a,

where F(x)=Rf (x) dx (withC=0). So finding a definite integral involves finding an indefinite integral. Sometimes the indefinite integral requires substitution. This section explains the process.

To motivate the topic, we will do our next example two ways. The first way is longer. The second, shorter solution highlights a preferred shortcut.

Example 45.12 FindZ1 0

°

x2+1¢52x dx.

Solution A Applying the fundamental theorem of calculus yields

Z1 0

°

x2+1¢52x dx = hF(x)i1

0 (§)

Before going further, we need to computeF(x)=R°x2₊₁¢5_{2x dx}_{. For this, let}

u=x2+1 , so du_dx=2x, and du=2x dx . Then F(x) = Z°x2+1¢52x dx =

Z

u5du = u₆6 =(x2+₆1)6.

We now get an answer by insertingF(x)=°x2₊₁¢6_/6_{into Equation (§), above.}

Z1 0 ° x2+1¢52x dx = h(x 2₊₁₎6 6 i1 0 = (12+1)6 6 ° (02+1)6 6 = 63 6 = 21 2 .

Solution B This time we’ll do the substitutions in-line, rather than

find-ingF(x)separately, as above. As before, let u=x2+1 , so du=2x dx . Then Z1 0 ° x2+1¢52x dx = Z_x=1 x=0 u 5_du

We have almost totally switched fromxtou. But the limits of integration are stillx-values. (In the original integral, xgoes from0to1.) We have em-phasized this by writing them asx=0andx=1. They must be converted tou. Sinceu=x2₊₁_, _x₌₀_gives_u₌₀2₊₁_{, and}_x₌₁_gives_u₌₁2₊₁_{. Our computation}

continues: = Z_u=12₊₁ u=02₊₁ u 5_du ₌ Z2 1 u 5_du₌∑u6 6 ∏2 1 = 26 6 ° 16 6 = 63 6 = 21 2 .

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We have now foundR1

0°x2+1¢52x dxin two different ways, getting the same

answer both times.

The moral of this example is that if we need to do a substitutionu=g(x) in a definite integral, then it is quicker to do it in-line, as in Solution B. However, in doing this we must remember to change the limits of integration aand bto g(a)and g(b).

We summarize this discussion by pairing the previous section’s substitu-tion rule (for indefinite integrals) with its companion for definite integrals.

Substitution Rule for Indefinite Integrals

Ifu=g(x), then Zf°g(x)¢g0_{(x) dx} ₌ Z_{f (u) du}_. Substitution Rule for Definite Integrals

Ifu=g(x), then Zb a f ° g(x)¢g0_{(x) dx} ₌ Zg(b) g(a) f (u) du. Example 45.13 FindZ2 1 5 (5x°1)2dx.

Solution Let u=5x°1 , so du_dx=5and du=5 dx . Then Z₂ 1 5 (5x°1)2dx= Z₂ 1(5x°1) °2_{5 dx}₌Z5·2°1 5·1°1 u °2_du₌Z9 4 u °2_du₌∑°1 u ∏9 4= 5 36. Example 45.14 FindZ0 °1 7 x2₊_2x₊₂dx.

Solution Perhaps your first impulse is to put u= x2+2x+2, but then du=(2x+1) dx is nowhere in sight. After some trial and error you may notice that this integral is similar toR 1

1+u2du=tan°1(u)+C. In fact,

Z₀ °1 7 x2₊_2x₊₂dx = 7 Z₀ °1 1 1+(x2₊_2x₊₁₎dx = 7 Z₀ °1 1 1+(x+1)2dx So let u=x+1 . Then du

dx=1, so du=dx . The computation continues as:

= 7Z0+1 °1+1 1 1+u2du = 7 h tan°1_(u)i1 0 = 7 ≥ º 4°0 ¥ = 7₄º.

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Example 45.15 FindZ1 0

x+1 x2₊_2x₊₂dx. Solution Let u=x2+2x+2 . Then du

dx=2x+2, so 12du=(x+1) dx . Then Z₁ 0 x+1 x2₊_2x₊₂dx = Z₁ 0 1 x2₊_2x₊₂(x+1) dx = Z12_+2·1+2 02_+2·0+2 1 u 1 2du = 1 2 Z5 2 1 udu = 1 2 h ln|u|i5₂ = 1₂°ln(5)°ln(2)¢ = 1₂lnµ5 2 ∂ = ln r 5 2.

Exercises for Chapter 45

Find the integrals.

1. Z2e°x_dx _2. Z_{4sin(3x) dx} 3. Z_2x4x₂ ° 6dx 4. Z 12s2p4s3_{+ 15 ds} 5. Zxp1 ° x2_dx _6. Z_sin(x)ecos(x)_dx 7. Zsin6(x)cos(x) dx _8. Z 1 x2 s 2 °1 x dx 9. Z6µ cos(3µ2) dµ 10. Z e2x24x dx 11. Z sin(2x)_cos₅_(2x)dx 12. Z 4x e2x2 dx 13. Z3 0 e °5x_dx _14. Z2 1 x +1 ° x2_{+ 2x}¢2 dx 15. Z1 0 x(x 2_{+ 1)}5_dx _16. Zº/6 °º/6tan(2x)sec(2x) dx 17. Z3 1 3x2_{+ 2x + 1} x3_{+ x}2_{+ x} dx 18. Z1 0 x p x2_{+ 1 dx}

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Exercise Solutions for Chapter 45 1. Letu = °xso du_dx = °1and°du = dx.

ThenR2e°x_{dx = 2}R_e°x_{dx = 2}R_eu_{(°du)2 = °2}R_eu_{du = °2e}u_{+ C = °2e}°x_{+ C}

3. Letu = 2x2° 6so du_dx = 4xanddu = 4x dx. Then

Z _4x 2x2_{° 6}dx = Z ₁ 2x2_{° 6}4x dx = Z 1 udu = ln|u|+ C = ln|2x2° 6| + C 5. Letu = 1° x2_{, so} du dx = °2x, hence°12du = x dx. Then Z xp1 ° x2_{dx =}Z°_{1 ° x}2¢1/2_{x dx =}Z_u1/2µ_°1 2du ∂ = °1₂ Z u1/2du = °1₂2₃u3/2+ C °pu 3 3 + C = ° p 1 ° x23 3 + C

7. Letu = sin(x)so du_dx= cos(x)anddu = cos(x) dxThen Z

sin6(x)cos(x) dx =Zu6du =u₇7+ C =sin

7_(x)

7 + C

9. Letu = 3µ2, so dudµ = 6µanddu = 6µ dµ. Then

Z

6µ cos(3µ2) dµ =Zcos(3µ2)6µ dµ =Zcos(u) du = sin(u)+ C = sin(3µ2) + C 11. Letu = cos(2x)so du

dx = °2sin(2x)and°12du = sin(2x) dx. Then

Z sin(2x) cos5_(2x)dx = Z (cos(2x))°5_{sin(2x) dx =}Z_u°5µ_°1 2du ∂ =1₂ Z u°5_{du =}1 2· u°4 °4 + C °_8u1₄+ C = °_8cos1₄_(2x)+ C 13. Letu = °5xso du dx = °5and°15du = dx. Then Z3 0 e °5x_{dx =}Z°5·3 °5·0 e uµ_°1 5du ∂ = °1₅ Z_°15 0 e u_{du = °}1 5 h eui°15 0 = ° 1 5 ° e°15_{° e}0¢₌1 5° 1 5e15

15. Letu = x2+ 1, so du_dx = 2xand 1₂du = x dx. Then Z1 0 x(x 2₊₁₎5_{dx =}Z1 0 (x 2₊₁₎5_{x dx =}Z1 2₊₁ 02₊₁ u 51 2du = 1 2 Z2 1 u 5_{du =}1 2 ∑ u6 6 ∏2 1= 1 2 µ 26 6° 16 6 ∂ =21₄ 17. Letu = x3_{+ x}2_{+ x}_{, so} du dx= 3x2+ 2x + 1, hencedu = ° 3x2_{+ 2x + 1}¢_dx_{. Then} Z3 1 3x2_{+ 2x + 1} x3_{+ x}2_{+ x} dx = Z3 1 1 x3_{+ x}2_{+ x} ° 3x2+ 2x + 1¢dx =Z3 3₊₃2₊₃ 13₊₁2₊₁ 1 udu = Z39 3 1 udu =hln|u|i39₃ = ln|39| ° ln|3| = lnµ 39₃ ∂ = ln(13)