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41 views8 pagesMath 210-Lecture 11 PDF
Uploaded by
Zeinab FarhatCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Chapter 8: Techniques of Integration
Section 8.3: Trigonometric Substitutions
Lecture 11
Outline
1- Procedure for Trigonometric Substitutions
2- Exercises
Course: Math 210 ‘Spring 2021-2022,
Dr. Sara Nasser seranasser@liuedulb Step 1-What do you see?? (01)
fomai(
@
Use the substitution
Ve —-x ——— | x=asind
IMPORTANT Step 2- Necessary Condition
Domain: —1
2 mee
Range: -%
y
y =sinx
Domain: [—7/2, 7/2]
Range: [—1, 1]
oS Step 3- Integrate
Step 4- Convert the Integral back to
the original variable x EXAMPLE { Evaluate
/
x=3sind, dx =3cos0db, —><0<5
Solution We set
9 — x? = 9 — 9sin?@ = 9(1 — sin? @) = 9cos* 0.
Then
_ [ 9sin? 6-3 cos 6 dO
[3 cos 6|
=
9 f sin? ao cos > Ofer —F <0 Ofor-—F
Vine
IMPORTANT Step 2- Necessary Condition
Domain: x =-lorx>1
Range: OS ySa,y# z
y = secx {
—__1_} x
Domain: [0, 7/2) U (77/2, 7] Boas fa
Range: (—00, —1] U[1, 00) a i.
S Step 3- Integrate
Step 4- Convert the Integral back to
the original variable x EXAMPLE 3 Evaluate
[Ss x>Z
V25x7 = 4 2
Solution We first rewrite the radical as
Vist =4 = \fas(e - 4)
to put the radicand in the form x? — a. We then substitute
2 ees =
x= Zsecd, dx 5 sec @ tan 6 dO, 0<0<5
2
2— (2) = Accra -+
x (2) 35 sec" 8 35
ay
= 3g (sec*@ — 1) = 55 tan’ @
2
2 2 2 tan > 0 for
ro (2) = § |tano| = Ztand. 0<0