Problem 1 (WA3 - 9).
7
Z
√
49 + x2 dx
0
solution. Let x = 7 tan θ, dx = 7 sec (θ) dθ; limits of integration 0 7→ 0, 7 7→ tan−1 (1) = π/4.
Thus
Z 7√
Z π/4 q
2
49 + x dx =
49 + 49 tan2 (θ)7 sec2 (θ) dθ
0
0
Z π/4 q
1 + tan2 (θ) sec2 (θ) dθ
= 49
0
Z π/4
sec3 θ dθ
= 49
0
π/4 1
= 49 ·
sec θ tan θ + ln | sec θ + tan θ| 2
0
√
√
49
=
2 + ln | 2 + 1| .
2
2
Problem 2 (WA3 - 28).
Z
dx
25x2 − 9
solution. Let x = 53 sec θ, dx = 53 sec θ tan θ dθ. The integral becomes
Z
Z
3
sec θ tan θ
1
5
dx
=
dθ
2
2
25x − 9
25 35 sec θ − 9
Z
3
sec θ tan θ
=
dθ
5
9 sec2 θ − 9
Z
1
sec θ tan θ
=
dθ
15
tan2 θ
Z
1
sec θ
=
dθ
15
tan θ
Z
1
=
csc θ dθ
15
1
= − ln | csc θ + cot θ| + C.
15
From here it remains to draw the triangle determined by your substitution and go back to
the original variable. Making diagrams is very time consuming in this markup language I
use to type math, so I’ll leave that part to you. If you get stuck on it, just send me an email
and I’ll scan a hand-drawing or something like that.
alternate solution. Notice that 25x12 −9 is a rational function, so we can decompose it using
the method of partial fractions!
1
25x2
−9
=
1
A
B
=
+
, so
(5x − 3)(5x + 3)
5x − 3 5x + 3
1 = A(5x + 3) + B(5x − 3) = (5A + 5B)x + (3A − 3B)
Next, as always with this method, we equate the coefficients of the powers of x on each side
of the equation. The coefficient of x1 on the left hand side of the above equation is 0, so we
have 0 = 5A + 5B. The coefficient of x0 on the left hand side is 1, so we have 1 = 3A − 3B.
Hence we have the system of linear equations
(
0 = 5A + 5B
1 = 3A − 3B
which we can solve using elimination or substitution. Either way, we find A = 16 and B = − 61 .
Hence our integral can be written
Z
Z Z 1
1
1
1
1
1
1
1
dx =
dx = · · ·
dx =
·
− ·
−
25x2 − 9
6 5x − 3 6 5x + 3
6
5x − 3 5x + 3
1
··· =
6
ln |5x − 3| ln |5x + 3|
−
5
5
1 5x − 3 +C
+C =
ln
30 5x + 3 where the last equality comes from properties of the logarithm and absolute value.
Both approaches are equally valid, and the answers are equivalent. You can transform
one into the other with careful algebra. However, looking at webassign’s answer key, I see
that they are expecting you to solve the problem using partial fractions.