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Final

# MA 141 Lecture 24: SubstitutionExam

Department
Mathematics
Course Code
MA 141
Professor
Kevin Flores
Study Guide
Final

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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

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Ex: Zxâˆš6âˆ’x2dx.
What happens if we choose something else for u? Could it still work?
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