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