Evaluate the integrals in Exercises 41–60.
43. ∫6cosh(x/2 - ln3)dx

Verified step by step guidance

1. Recall the definition of the hyperbolic cosine function: \(\cosh(u) = \frac{e^{u} + e^{-u}}{2}\). This can help in rewriting the integral if needed.
2. Identify the inner function inside the hyperbolic cosine: here, \(u = \frac{x}{2} - \ln 3\). This will be useful for substitution.
3. Use substitution by letting \(u = \frac{x}{2} - \ln 3\). Then, compute \(\frac{du}{dx} = \frac{1}{2}\), which implies \(dx = 2 \, du\).
4. Rewrite the integral in terms of \(u\): \(\int 6 \cosh\left(\frac{x}{2} - \ln 3\right) dx = \int 6 \cosh(u) \cdot 2 \, du = \int 12 \cosh(u) \, du\).
5. Integrate \(\cosh(u)\) with respect to \(u\): \(\int \cosh(u) \, du = \sinh(u) + C\). So, the integral becomes \(12 \sinh(u) + C\). Finally, substitute back \(u = \frac{x}{2} - \ln 3\) to express the answer in terms of \(x\).

Key Concepts

Hyperbolic Cosine Function (cosh)

The hyperbolic cosine function, cosh(x), is defined as (e^x + e^(-x))/2. It is an even function and appears frequently in calculus problems involving hyperbolic functions. Understanding its properties and derivatives is essential for integrating expressions involving cosh.

Integration of Composite Functions

When integrating functions like cosh(ax + b), it is important to use substitution or recognize the integral form. The integral of cosh(u) with respect to u is sinh(u), so adjusting for the inner function's derivative is necessary to find the correct antiderivative.

Logarithmic Constants in Integration

Constants such as ln(3) inside the argument of a function affect the integration as shifts but do not change the integral's form. Recognizing that ln(3) is a constant helps simplify the integral by treating it as a constant shift in the variable.