Calculus- Early Transcendentals, 2021a

7.3. Trigonometric Substitutions 273
Solution. First, complete the square to write
3 − 2x − x 2 = 4 − (x + 1) 2
Now, we may let u = x + 1sothatdu = dx (note that x = u − 1) to get:
∫
∫
x
√ dx = 4 − (x + 1) 2
u − 1
√
4 − u 2 du
Let u = 2sinθ giving du = 2cosθ dθ:
∫
∫ ∫
u − 1 2sinθ − 1
√ du = · 2cosθ dθ = (2sinθ − 1)dθ
4 − u 2 2cosθ
Integrating and using a triangle we get:
∫
x
√
3 − 2x − x 2
= −2cosθ − θ +C
= −
√4 ( − u 2 − sin −1 u
)
+C
2 ( ) x + 1
= −
√3 − 2x − x 2 − sin −1 +C
2
Note that in this problem we could have skipped the u-substitution if instead we let x + 1 = 2sinθ. (For
the triangle we would then use sinθ = x + 1
2 .) ♣
Exercises for 7.3
Exercise 7.3.1
Exercise 7.3.2
Exercise 7.3.3
Exercise 7.3.4
Exercise 7.3.5
∫ √
x 2 − 1dx
Exercise 7.3.6
∫ ∫
√
9 + 4x 2 dx
Exercise 7.3.7
∫ √
∫
x 1 − x 2 dx
Exercise 7.3.8
∫
x 2√ ∫
1 − x 2 dx
Exercise 7.3.9
∫
∫
1
√ dx Exercise 7.3.10
1 + x 2
∫ √
x 2 + 2xdx
1
x 2 (1 + x 2 ) dx
x 2
√
4 − x 2 dx
√ x
√
1 − x
dx
x 3
√
4x 2 − 1 dx
Apply Trigonometric Substitution to evaluate the indefinite integrals.