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**preview**shows pages 1-2. to view the full**6 pages of the document.**MA 141 Chapter 4

Section 4.4: Substitution

As we saw previously, sometimes integrals represent a function composition. Substitution is

a method to deal with such integrals. Recall that if H(x) = F(g(x)) and Fâ€²=f, then

Hâ€²(x) = Fâ€²(g(x)) Â·gâ€²(x) = f(g(x)) Â·gâ€²(x)

Zf(g(x)) Â·gâ€²(x)dx =F(g(x)) + C.

We can simplify the problem by letting u=g(x). Recall that we said in Section 3.6 if

y=f(x), then dy =fâ€²(x)dx. Then here we have du =gâ€²(x)dx and

Zf(g(x)) Â·gâ€²(x)dx =Zf(u)du =F(u) + C=F(g(x)) + C.

Using this change of variables is called the substitution method. For substitution to work,

we pick part of our integrand to be uand ensure that this relabeling with uand du leaves

behind none of the old variable (the change must be complete).

Ex: Find Zxâˆš6âˆ’x2dx.

Letâ€™s try using u= 6âˆ’x2. Then du =âˆ’2xdx. How does this relabel our integral? We cannot

relabel just yet because we do not have the full du term. We are missing the coeï¬ƒcient âˆ’2.

Zxâˆš6âˆ’x2dx =âˆ’1

2Zâˆ’2xâˆš6âˆ’x2dx

=âˆ’1

2Zâˆšudu

=âˆ’1

2Zu1/2du

=âˆ’1

2Â·u3/2

3/2+C

=âˆ’1

3u3/2+C

We always want to go back to our original variable. Since u= 6 âˆ’x2, we have

Zxâˆš6âˆ’x2dx =âˆ’1

3u3/2+C=âˆ’1

3(6 âˆ’x2)3/2+C

Note: we could also solve for x dx and substitute that in to the integral. x dx =âˆ’du

2.

1

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

Ex: Zxâˆš6âˆ’x2dx.

What happens if we choose something else for u? Could it still work?

Try u=âˆš6âˆ’x2instead. Then

du =1

2(6 âˆ’x2)âˆ’1/2(âˆ’2x)dx

=âˆ’xdx

âˆš6âˆ’x2

âˆ’âˆš6âˆ’x2du =xdx

âˆ’u du =x dx

Zxâˆš6âˆ’x2dx =Zu(âˆ’u du)

=âˆ’Zu2du

=âˆ’1

3u3+C

=âˆ’1

3(6 âˆ’x2)3/2+C

Ex: Find Zsin(x) cos(x)dx.

What should I choose for u? Try u= sin(x),du = cos(x)dx.

Zsin(x) cos(x)dx =Zu du

=u2

2+C

=sin2x

2+C

Can I use u= cos x? Then du =âˆ’sin xdx.

Zsin(x) cos(x)dx =âˆ’Zcos x(âˆ’sin(x))dx

=âˆ’Zu du

=âˆ’u2

2+C

=âˆ’cos2x

2+C

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