Calculus 2nd Edition Rogawski

SECTION 8.3 Trigonometric Substitution 429
The logarithmic term can be rewritten as
(√ )
x
5ln
2 + 5 + x
(√ )
√ + C = 5ln x 2 + 5 + x − 5ln √ 5 + C
5 } {{ }
Constant
Since the constant C is arbitrary, we can absorb −5ln √ 5 into C and write
∫ √ (√ )
4x 2 + 20 dx = x√
x 2 + 5 + 5ln x 2 + 5 + x + C
Our last trigonometric substitution x = a sec θ transforms √ x 2 − a 2 into a tan θ because
x 2 − a 2 = a 2 sec 2 θ − a 2 = a 2 (sec 2 θ − 1) = a 2 tan 2 θ
In the substitution x = a sec θ, we choose
0 ≤ θ < π 2 if x ≥ a and π ≤ θ < 3π 2 if
x ≤−a. With these choices, a tan θ is the
positive square root √ x 2 − a 2 .
Integrals Involving √ x 2 − a 2
substitution
x = a sec θ,
If √ x 2 − a 2 occurs in an integral where a>0, try the
dx = a sec θ tan θ dθ,
√
x 2 − a 2 = a tan θ
∫
EXAMPLE 4 Evaluate
dx
x 2√ x 2 − 9 .
Solution In this case, make the substitution
∫
x = 3 sec θ,
∫
dx
x 2√ x 2 − 9 =
dx = 3 sec θ tan θ dθ,
3 sec θ tan θ dθ
(9 sec 2 θ)(3 tan θ) = 1 ∫
9
Since x = 3 sec θ, we use the right triangle in Figure 3:
√
x 2 − 9 = 3 tan θ
cos θ dθ = 1 9 sin θ + C
FIGURE 3
θ
x
3
x 2 − 9
Therefore,
sec θ = hypotenuse
adjacent
∫
= x √
opposite x
, sin θ =
3 hypotenuse = 2 − 9
x
dx
x 2√ x 2 − 9 = 1 √
x
9 sin θ + C = 2 − 9
+ C
9x
So far we have dealt with the expressions √ x 2 ± a 2 and √ a 2 − x 2 . By completing
the square (Section 1.2), we can treat the more general form √ ax 2 + bx + c.
∫
EXAMPLE 5 Completing the Square Evaluate
Solution
Step 1. Complete the square.
dx
(x 2 − 6x + 11) 2 .
x 2 − 6x + 11 = (x 2 − 6x + 9) + 2 = (x − 3)
} {{ }
2 +2
u 2