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Unformatted text preview: Lecture 5: MATH 138W12003 January 13, 2012 I) Trigonometric substitution, Section 7.3
√
√
In this lecture we focus on computing integrals involving functions that include a2 − x2 , x2 − a2
√
and x2 + a2 using trigonometric substitutions. For example the ﬁrst occurs when we want to
compute the area of an ellipse.
Example If the equation of an ellipse is x2 /a2 + y 2 /b2 = 1 then we can write the are as,
a A=4
0 b√ 2
a − x2 dx.
a At the moment we do not have the techniques for evaluating this. The new technique that we begin to develop is to make a trigonometric substitution of the form, x = a sin θ and dx = a cos θdθ.
The textbook calls this an inverse substitution because we are making this more complicated but
it is still a substitution so that’s the terminology I’m going to use.
There is a geometric interpretation of this substitution. We are deﬁning a triangle that has one
√
side x, another a2 − x2 and a hypotenuse of a. We can assume that the angle we are dealing
with is in the range −π/2 ≤ θ ≤ π/2. If x is positive than so is the angle but if x is negative, then
the angle is negative. It is important to recall that cos θ is not negative in this interval so that
 cos θ = cos θ.
To evaluate the integral we substitute our change of variables into the integral to yield after
using our trigonometric substitution,
a b
a A=4
0 a2 − a2 sin2 θ a cos θdθ, π /2 1 − sin2 θ cos θdθ, = 4ab
0 sub in trig identity π /2  cos θ cos θdθ, = 4ab
0 but cos is positive π /2 cos2 θdθ. = 4ab
0 This integral is something we know how to solve by using a double angle substitution,
π /2 A = 2ab
0 1
(1 + cos 2θ)dθ = 2ab θ + sin 2θ
2 π /2 = πab
0 Note that in the special case where a = b then we recover the formula for the area of a circle.
The strategy that we should use is the following depending on the form of the square root.
√
1) If we have a2 − x2 then we make the substitution,
x = a sin θ, − π
π
≤θ≤ ,
2
2 and we need to use the identity 1 − sin2 θ = cos2 θ. The square root always takes the form
√
a2 − x2 = a cos θ = a cos θ.
1 2) If we have √
a2 + x2 then we make the substitution,
− x = a tan θ, π
π
≤θ≤ ,
2
2 and we need to use the identity 1 + tan2 θ = sec2 θ. The square root always takes the form
√
a2 + x2 = a sec θ = a sec θ.
√
3) If we have x2 − a2 then we make the substitution,
x = a sec θ, 0≤θ≤ π
2 or π≤θ≤ 3π
,
2 and we need to use the identity sec2 θ − 1 = tan2 θ. In the above domain, the ﬁrst is for x
√
positive and the second for x negative. The square root always takes the form x2 − a2 =
a tan θ = a tan θ. a x x
sin θ = ,
a x sin θ = √ cos θ = √ a2 − x2
,
a tan θ = √ θ
a2 − x2
a2 + x2
θ a2 x
,
+ x2 cos θ = √ x2 − a2
,
x a
cos θ = ,
x a2 a
,
+ x2 x
a2 − x2 tan θ = x
a a
x
θ x2 − a2 sin θ = √ tan θ = √ x2 − a2
x a Figure 1: In this Figure we see the three different types of triangles that we have depending on
which case we need to consider.
Example: Find √xx+4 dx.
2
Before we jump into using a trigonometric substitution we should stop and think about the
integrand. In the denominator we have the square root of a quadratic. The derivative of a
quadratic is proportional to x, which is what appears in the numerator. Therefore, there are two
methods at our disposal.
2 Method 1 The simplest option is to substitute u = x2 + 4, du = 2xdx, which yields,
√
√
x
1
du
√
√ = u + C = x2 + 4 + C.
dx =
2
u
x2 + 4
Method 2 A trigonometric substitution that√
works for this problem is x = 2 tan θ and dx =
2
2 sec θdθ. Note that this implies that cos θ = 2/ x2 + 4. From this we can evaluate the integral
in question,
2 tan θ
√
2 sec2 θdθ,
2
4 tan θ + 4
tan θ
=2 √ 2
sec2 θdθ, now use trig identity
tan θ + 1
tan θ
=2
sec2 θdθ, but cos is positive
 sec θ x
√
dx =
x2 + 4 =2 tan θ sec θdθ, this is a standard integral = 2 sec θ + C, substitute in for θ,
√
= x2 + 4 + C.
Of course the two methods yield the same answer, as they must.
Example: Evaluate √xdx a2 for a > 0. We begin with the substitution, x = a sec θ, dx =
2−
a sec θ tan θdθ on the intervals 0 < θ < π/2 or π < θ < 3π/2. In either of these intervals the square
root can be reduced as,
√
√
√
√
x2 − a2 = a2 sec2 θ − a2 = a sec2 θ − 1 = a tan2 θ = a tan θ = a tan θ.
We can substitute this into the integral to obtain,
dx
√
=
x 2 − a2
= a sec θ tan θ
dθ,
a tan θ
sec θdθ, = ln  sec θ + tan θ + C.
The last lines arises from a formula that we derived in Lecture 3. Given the triangle we can now
rewrite this in terms of x,
√
√
x
x2 − a2
dx
√
= ln  +
 + C = ln x + x2 − a2  + C.
a
a
x 2 − a2 In the above we used the property ln(ab) = ln a + ln b and absorbed − ln a into √ constant.
the
Question: What do we do if we have an integrand that has a function like ax2 + bx + c?
Solution: Compute the square,
√ ax2 + bx + c =
= b
a(x2 + x) + c,
a
b
b2
a(x2 + )2 + c − .
2a
4a Then depending on the sign of c − b2 /(4a) we can use one of our previously mentioned methods.
3 ...
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This note was uploaded on 02/14/2012 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.
 Fall '07
 Anoymous
 Calculus, Integrals

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