Math Problem Statement

Evaluate the integral ∫ 2 cosh(t) [sech(t) + csch²(t)] dt.

Solution

To solve the integral

2cosht[secht+csch2t]dt,\int 2 \cosh t \left[ \text{sech} \, t + \text{csch}^2 \, t \right] dt,

we proceed as follows:

Step 1: Distribute the terms

Expand the product inside the integral: 2cosht[secht+csch2t]dt=2coshtsechtdt+2coshtcsch2tdt.\int 2 \cosh t \left[ \text{sech} \, t + \text{csch}^2 \, t \right] dt = \int 2 \cosh t \, \text{sech} \, t \, dt + \int 2 \cosh t \, \text{csch}^2 \, t \, dt.

Step 2: Simplify the first term

Using the identity coshtsecht=1\cosh t \cdot \text{sech} \, t = 1: 2coshtsechtdt=2dt=2t.\int 2 \cosh t \, \text{sech} \, t \, dt = \int 2 \, dt = 2t.

Step 3: Simplify the second term

Recall the derivative of cotht\text{coth} \, t: ddtcotht=csch2t.\frac{d}{dt} \text{coth} \, t = -\text{csch}^2 \, t. Thus: 2coshtcsch2tdt=2coshtddt(cotht)dt.\int 2 \cosh t \, \text{csch}^2 \, t \, dt = -2 \int \cosh t \cdot \frac{d}{dt} (\text{coth} \, t) \, dt. This simplifies to: 2coshtcotht.-2 \cosh t \, \text{coth} \, t.

Step 4: Combine results

Combining the results from both parts: 2cosht[secht+csch2t]dt=2t2coshtcotht+C,\int 2 \cosh t \left[ \text{sech} \, t + \text{csch}^2 \, t \right] dt = 2t - 2 \cosh t \, \text{coth} \, t + C, where CC is the constant of integration.

Would you like further clarification or details on any of the steps? Here are some related questions for further exploration:

  1. What are the key hyperbolic function identities used in integrals?
  2. How does the derivative of cotht\text{coth} \, t relate to this integral?
  3. Can substitution techniques simplify integrals involving hyperbolic functions further?
  4. What are practical applications of hyperbolic integrals in physics or engineering?
  5. How does the integration process differ for definite integrals of similar functions?

Tip: Memorize basic hyperbolic function derivatives and identities—they often simplify complex integrals.

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Math Problem Analysis

Mathematical Concepts

Hyperbolic Functions
Integral Calculus
Simplification of Expressions

Formulas

cosh(t) · sech(t) = 1
Derivative of coth(t): d/dt(coth(t)) = -csch²(t)

Theorems

Properties of Hyperbolic Functions
Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate (Calculus II or III)