James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

488 Chapter 7 Techniques of Integration
the limits of integration we note that when x − 0, sin − 0, so − 0; when x − a,
sin − 1, so − y2. Also
sa 2 2 x 2 − sa 2 2 a 2 sin 2 − sa 2 cos 2 − a | cos | − a cos
since 0 < < y2. Therefore
A − 4 b a ya sa 2 2 x 2 dx − 4 b
0 a yy2 a cos ? a cos d
0
− 4ab y y2
cos 2 d − 4ab y y2 1
2 s1 1 cos 2d d
0
− 2abf 1 1 2 sin 2g 0
0
y2
− 2ab S 2 1 0 2 0 D − ab
We have shown that the area of an ellipse with semiaxes a and b is ab. In particular,
taking a − b − r, we have proved the famous formula that the area of a circle with
radius r is r 2 .
NOTE Since the integral in Example 2 was a definite integral, we changed the limits
of integration and did not have to convert back to the original variable x.
Example 3 Find y
1
x 2 sx 2 1 4
dx.
SOLUTION Let x − 2 tan , 2y2 , , y2. Then dx − 2 sec 2 d and
sx 2 1 4 − s4stan 2 1 1d − s4 sec 2 − 2 | sec | − 2 sec
n
So we have
y
dx
x 2 sx 2 1 4
− y
2 sec 2 d
4 tan 2 ? 2 sec − 1 y sec
4 tan 2 d
To evaluate this trigonometric integral we put everything in terms of sin and cos :
sec
tan 2 − 1
cos ? cos2
sin 2 − cos
sin 2
Therefore, making the substitution u − sin , we have
y
dx
− 1 y cos
x 2 sx 2 1 4 4 sin 2 d − 1 y du
4 u 2
¨
œ„„„„„ ≈+4
2
x
− 1 4
S2 1 uD 1 C − 2 1
4 sin 1 C
− 2 csc
4
1 C
We use Figure 3 to determine that csc − sx 2 1 4 yx and so
FIGURE 3
tan − x 2
y
dx
− 2 sx 2 1 4
x 2 sx 2 1 4 4x
1 C n
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