MA1S12: SOLUTIONS TO TUTORIAL 2
1. Use a trigonometric substitution to compute the indefinite integral
Z
√
3
dx
4 + x2
Solution: We use the trigonometric substitution
x = 2 tan θ
=⇒ dx =
dx
dθ = 2 sec2 θ dθ
dθ
We also have
√
4 + x2 =
p
p
4 + 4 tan2 θ = 2 1 + tan2 θ = 2 sec θ
and so
Z
3
√
dx = 3
4 + x2
Z
1
(2 sec2 θ) dθ
2 sec θ
= 3
Z
sec θ dθ
= 3 ln | sec θ + tan θ|
√
4 + x2 x + + const
= 3 ln 2
2
2. Compute the area of the region which is bounded by the graph of
√
f (x) =
between 0 and 4.
1
16 − x2
2
2
MA1S12: SOLUTIONS TO TUTORIAL 2
Region bounded by the graph of f HxL between 0 and 4
2.0
1.5
1.0
0.5
1
2
3
4
Solution: The area is given by the definite integral
Z
4
0
√
16 − x2
dx
2
We apply a trigonometric substitution
x = 4 sin θ
=⇒ dx =
dx
dθ = 4 cos θ dθ
dθ
Note that
√
16 − x2 =
p
p
16 − 16 sin2 θ = 4 1 − sin2 θ = 4 cos θ
To find an antiderivative we compute the indefinite integral,
Z √
16 − x2
dx =
2
Z
8 cos2 θ dθ
1
θ
= 8
cos θ sin θ +
+ const
2
2
√
16 − x2 x 1
−1 x
= 8
+ sin
+ const
8
4 2
4
MA1S12: SOLUTIONS TO TUTORIAL 2
3
Now the area is
4
Z
0
√
16 − x2
dx = 8
2
√
x
16 − x2 x 1
+ sin−1
8
4 2
4
i
h
π
− 8 [0]
= 8 0+
4
4
0
= 2π
3. Use the method of partial fractions to compute the indefinite integral
Z
x2
dx
(x − 1)(x2 + 2x + 1)
Solution: Factorizing the denominator gives
x2
x2
=
(x − 1)(x2 + 2x + 1)
(x − 1)(x + 1)2
Now we write
x2
A
B1
B2
=
+
+
(x − 1)(x2 + 2x + 1)
x − 1 x + 1 (x + 1)2
Multiplying across by the denominator gives
x2 = A(x + 1)2 + B1 (x − 1)(x + 1) + B2 (x − 1)
= A(x2 + 2x + 1) + B1 (x2 − 1) + B2 (x − 1)
= (A + B1 )x2 + (2A + B2 )x + (A − B1 − B2 )
Equating coefficients of powers of x we get a system of linear equations,
A + B1 = 1
2A + B2 = 0
A − B1 − B2 = 0
4
MA1S12: SOLUTIONS TO TUTORIAL 2
Solving we find
1
A= ,
4
3
B1 = ,
4
B2 = −
1
2
Thus we have a partial fraction decomposition
x2
1
3
1
=
+
−
2
(x − 1)(x + 2x + 1)
4(x − 1) 4(x + 1) 2(x + 1)2
Now
Z
x2
dx =
(x − 1)(x2 + 2x + 1)
Z
Z
Z
1
3
1
dx +
dx −
dx
4(x − 1)
4(x + 1)
2(x + 1)2
1
3
1
=
ln |x − 1| + ln |x + 1| +
+ const
4
4
2(x + 1)
4. Use the method of partial fractions to compute the indefinite integral
Z
3x2 + x + 4
dx
x3 + x
Solution: First we factorize the denominator
x3 + x = x(x2 + 1)
We have a linear factor x and a quadratic factor x2 + 1. Next we
write
3x2 + x + 4
A Bx + C
=
+ 2
x3 + x
x
x +1
To find A, B and C we multiply across by the denominator x3 + x
to get
3x2 + x + 4 = A(x2 + 1) + (Bx + C)(x)
= (A + B)x2 + Cx + A
MA1S12: SOLUTIONS TO TUTORIAL 2
5
Equating coefficients of powers of x we get a system of linear equations
A+B = 3
C = 1
A = 4
Solving we find B = −1. Thus we have a partial fraction decomposition
3x2 + x + 4
4 −x + 1
= + 2
3
x +x
x
x +1
The integral becomes
Z
3x2 + x + 4
dx =
x3 + x
Z
4
−x + 1
dx +
dx
x
x2 + 1
Z
Z
Z
1
x
1
dx +
dx
= 4
dx −
2
2
x
x +1
x +1
1
= 4 ln |x| − ln |x2 + 1| + tan−1 x + const
2
Z