Calculus Volume 1

Chapter 6 | Applications of Integration

747

d dx (cosh

x ) 2

b.

Solution Using the formulas in Table 6.2 and the chain rule, we get a. d dx ⎛ ⎝ sinh ⎛ ⎝ x 2 ⎞ ⎠ ⎞ ⎠ =cosh ⎛ ⎝ x 2 ⎞ ⎠ ·2 x

d dx (cosh

x ) 2 =2cosh x sinh x

b.

6.47

Evaluate the following derivatives:

⎛ ⎝ tanh ⎛ ⎝ ⎜ 1

⎛ ⎝ x 2 +3 x ⎞ ⎠ ⎞ ⎠

d dx d dx

a.

⎞ ⎠ ⎟

b.

(sinh x ) 2

Example 6.48 Integrals Involving Hyperbolic Functions

Evaluate the following integrals: a. ∫ x cosh ⎛ ⎝ x 2 ⎞ ⎠ dx b. ∫ tanh xdx

Solution We can use u -substitution in both cases.

a. Let u = x 2 . Then, du =2 xdx and ∫ x cosh ⎛ ⎝ x 2 ⎞ b. Let u =cosh x . Then, du = sinh xdx and ∫ tanh xdx = ∫ sinh x cosh x ⎠ dx = ∫ 1

2 sinh ⎛

⎞ ⎠ + C .

⎝ x 2

udu = 1

u + C = 1

2 cosh

2 sinh

dx = ∫ 1

u du = ln| u | + C = ln | cosh x | + C .

Note that cosh x >0 for all x , so we can eliminate the absolute value signs and obtain ∫ tanh xdx = ln(cosh x )+ C .

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