Calculus Volume 1

Chapter 6 | Applications of Integration

751

Solution Using the formulas in Table 6.4 and the chain rule, we obtain the following results: a. d dx ⎛ ⎝ sinh −1 ⎛ ⎝ x 3 ⎞ ⎠ ⎞ ⎠ = 1 2 = 1

9+ x 2

3 1+ x

9

⎛ ⎝ tanh −1 x ⎞ ⎠

2

2

⎛ ⎝ tanh −1 x ⎞ ⎠

d dx

=

b.

1− x 2

6.49

Evaluate the following derivatives:

⎛ ⎝ cosh −1 (3 x ) ⎞ ⎠

d dx d dx

a.

3

⎛ ⎝ coth −1 x ⎞ ⎠

b.

Example 6.50 Integrals Involving Inverse Hyperbolic Functions

Evaluate the following integrals: a. ∫ 1 4 x 2 −1 dx b. ∫ 1 2 x 1−9 x 2 dx

Solution We can use u -substitution in both cases. a. Let u =2 x . Then, du =2 dx and we have ∫ 1 4 x 2 −1 dx = ∫ 1 2 u 2 −1 b. Let u =3 x . Then, du =3 dx and we obtain ∫ 1 2 x 1−9 x 2 dx = 1 2 ∫ 1 u 1− u 2

−1 u + C = 1

−1 (2 x )+ C .

du = 1

2 cosh

2 cosh

−1 | u | + C = − 1

−1 |3 x | + C .

du = − 1

2 sech

2 sech

6.50

Evaluate the following integrals: a. ∫ 1 x 2 −4 dx , x >2 b. ∫ 1 1− e 2 x dx

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