2001 Test 3 solution

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10 Apr 2012
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MAT 137Y Term Test 3 Solutions, 2001-2002
1. Evaluate the following integrals.
(6%) (i)
et
1
etdt.
Let u
1
et. Then du
etdt, therefore giving us
et
1
etdt
u du
2
3u3
2
2
3
1
et
3
2
C
(6%) (ii)
sin5x
cos2xdx.
While this can be integrated by parts, it is easier to first rewrite the integral:
I
sin5x
cos2xdx
sin4x
cos2x
sin xdx
1
cos2x
2
cos2x
sin xdx
Now let u
cos x. Then du

sin xdx, giving us
I
1
u2
2
u2du
1
2u2
u4
u2du
1
u2
2
u2du
1
u
2u
u3
3
1
cos x
2 cos x
cos3x
3
C
(7%) (iii)
x2
x
1
x3
2x2
xdx.
We decompose the rational function using partial fractions:
x2
x
1
x3
2x2
x
x2
x
1
x
x
1
2
A
x
B
x
1
C
x
1
2
where A,B,Care constants. Solving for these constants:
A
x
1
2
Bx
x
1
Cx
x2
x
1

A
B
x2
2A
B
C
x
A
x2
x
1
Matching coefficients, we have A
1, A
B
1, and
2A
B
C

1. Solving, we have
B
0 and C
1. Hence,
x2
x
1
x3
2x2
xdx
1
x
1
x
1
2dx
ln
x

1
x
1
C
(7%) (iv)

3
1
1
dx
x2
2x
2
2.
The integral cannot be decomposed using partial fractions; the expression is already simplified
in partial fractions. Completing the square of the quadratic in the numerator, we have

3
1
1
dx
x2
2x
2
2

3
1
1
dx
x
1
2
1
2
Let x
1
tan θ. Then dx
sec2θdθ, so
3
1
1
dx
x
1
2
1
2
π
3
0
sec2θ
sec4θdθ
π
3
0
cos2θdθ
π
3
0
1
cos 2θ
2dθ
1
2θ
1
4sin 2θ
π
3
0
π
6
1
4sin 2π
3
π
6
1
4
3
2
π
6
3
8
1
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