Question

Evaluate the integral. \(\int_{0}^{\ln 2} 2 e^{-x} \cosh x d x\)

Short answer

The value of the integral is \( \ln2 +\frac{3}{8} \).

Step-by-step solution

Identify function form

The given integral, \( \int_{0}^{\ln 2} 2e^{-x} \cosh x dx \), involves the product of an exponential function and a hyperbolic cosine function.

Recognize the hyperbolic cosine

The hyperbolic cosine function \( \cosh x \) can be expressed in terms of exponential functions as \( \cosh x = \frac{e^{x} + e^{-x}}{2} \)

Substitute and simplify

Substitute \(\cosh x\) in this form into the integral. The result is \( \int_{0}^{\ln 2} 2e^{-x} \left(\frac{e^{x} + e^{-x}}{02}\right) dx \therefore \int_{0}^{\ln 2} e^{-x +x} + e^{-x - x} dx \), which simplifies to \( \int_{0}^{\ln 2} e^{0} + e^{-2x} dx = \int_{0}^{\ln 2} 1 + e^{-2x} dx \)

Compute the integral

To compute this integral, use the power rule for the first term and the rule for integrating an exponential function for the second term: \( \left[ x -\frac{1}{2}e^{-2x} \right]_{0}^{\ln 2} \)

Evaluate the Definite Integral

Substitute the values of the limits \( \ln2 \) and \(0\) to evaluate the integral: \( \left[ (\ln2 -\frac{1}{2}e^{-2 \ln 2} ) - (0 -\frac{1}{2}e^{0}) \right] = \ln2 -\frac{1}{2}e^{-2 \ln 2} +\frac{1}{2} \)

Simplify Result

Simplify the result \( \ln2 -\frac{1}{2}e^{-2 \ln 2} +\frac{1}{2}= \ln2 -\frac{1}{2} \cdot \frac{1}{2^2} +\frac{1}{2}= \ln2 -\frac{1}{8} +\frac{1}{2}= \ln2 +\frac{3}{8}\)