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Thomas Calculus 13th [Solutions]

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554 Chapter 8 Techniques of Integration<br />

45. Since secant is an even function and the domain is symmetric to 0, x 0.<br />

For the y-coordinate:<br />

y<br />

1<br />

2<br />

/4 2 1<br />

/4<br />

sec x dx<br />

/4 2<br />

tan x<br />

/4<br />

/4 /4<br />

sec x dx ln sec x tan x<br />

/4<br />

/4<br />

1<br />

ln 2 1 ln 2 1<br />

ln<br />

1 1<br />

2 1 ln 3 2 2<br />

2 1<br />

0.567<br />

46. Since both cosecant and the domain are symmetric around / 2, x / 2.<br />

y<br />

1 5 /6 2 1<br />

5 /6<br />

csc x dx<br />

/6 2<br />

cot x<br />

2<br />

/6<br />

5 /6 5 /6<br />

csc x dx ln csc x cot x<br />

/6<br />

/6<br />

1 3 3<br />

2<br />

ln 2 3 ln 2 3<br />

3 3<br />

0.658<br />

2 3 ln 7 4 3<br />

ln<br />

2 3<br />

47.<br />

48.<br />

3 3<br />

3 x x<br />

1 3x e dx xe C<br />

1<br />

dx<br />

2<br />

1 sin x<br />

Multiply the integrand by<br />

sec<br />

sec<br />

2<br />

2<br />

x<br />

.<br />

x<br />

2 2<br />

1 sec x<br />

sec x<br />

dx dx dx<br />

2 2 2 2<br />

1 sin x sec x tan x 1 2 tan x<br />

u tan x,<br />

du sec x dx<br />

2<br />

sec x<br />

1<br />

dx<br />

du<br />

2 2<br />

1 2 tan x 1 2u<br />

2<br />

v 2 u, dv 2 du<br />

1 1 1<br />

du<br />

1 2u<br />

2 1 v<br />

1 tan<br />

1<br />

v<br />

2<br />

2 2<br />

1 tan<br />

1 2 tan<br />

2<br />

dv<br />

C<br />

x<br />

C<br />

Copyright<br />

2014 Pearson Education, Inc.

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