TRIGONOMETRIC SUBSTITUTIONS We showed in class that the area enclosed by the circle x 2 + y 2 = a 2 equals πa 2 . It involved computing integrals of the form Z a 0 p a 2 - x 2 dx Exercise 1. Find the area enclosed by the ellipse x 2 a 2 + y 2 b 2 =1 The following table is a list of trigonometric substitutions that we use for certain radical expressions. Table 1. Table of Trigonometric Substitutions Type Expression Substitution Identity 1 √ a 2 - x 2 x = a sin θ, - π 2 ≤ θ ≤ π 2 1 - sin 2 θ = cos 2 θ 2 √ a 2 + x 2 x = a tan θ, - π 2 <θ< π 2 1 + tan 2 θ = sec 2 θ 3 √ x 2 - a 2 x = a sec θ, 0 ≤ θ< π 2 or π ≤ θ< 3π 2 sec 2 θ - 1 = tan 2 θ 1
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TRIGONOMETRIC SUBSTITUTIONS - McMaster Universityppoudel/Teaching/Fall-16/Homework/Lecture 32... · 2 TRIGONOMETRIC SUBSTITUTIONS Exercise 2. (Type 1) Z p 4 x2 x2 dx Note that the
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TRIGONOMETRIC SUBSTITUTIONS
We showed in class that the area enclosed by the circle x2 + y2 = a2 equals πa2. It involved
computing integrals of the form ∫ a
0
√a2 − x2 dx
Exercise 1. Find the area enclosed by the ellipse
x2
a2+y2
b2= 1
The following table is a list of trigonometric substitutions that we use for certain radical
expressions.
Table 1. Table of Trigonometric Substitutions
Type Expression Substitution Identity
1√a2 − x2 x = a sin θ, −π
2≤ θ ≤ π
21− sin2 θ = cos2 θ
2√a2 + x2 x = a tan θ, −π
2< θ <
π
21 + tan2 θ = sec2 θ
3√x2 − a2 x = a sec θ, 0 ≤ θ < π
2or π ≤ θ < 3π
2sec2 θ − 1 = tan2 θ
1
2 TRIGONOMETRIC SUBSTITUTIONS
Exercise 2. (Type 1)
∫ √4− x2x2
dx
Note that the radical term is of Type 1 i.e. radical of the form√
22 − x2.Therefore we need to use the substitution x = 2 sin θ, and proceed from there.
Exercise 3. (Type 2)
∫1
x2√x2 + 4
dx
Just because it has a radical expression does not necessarily mean that you need to use the
substitution from the table. For instance:
Exercise 4.
∫x√
x2 + 9dx
TRIGONOMETRIC SUBSTITUTIONS 3
Exercise 5. (Type 3)
∫1√
x2 − 1dx
So there are integrals where you need to do substitution first before computing the integrals
Exercise 6.
∫ √3
2
0
x3
(4x2 + 1)32
dx
4 TRIGONOMETRIC SUBSTITUTIONS
There are certain integrals where you need to solve by completing the squares first