Question

Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}} d x$$

Short answer

To evaluate the given integral, we first simplified the integrand by factoring out \(e^{3x}\) and cancelling it from the numerator and denominator. Then, we applied the substitution method with \(u = e^{-6x}\) and changed the limits accordingly. We found the antiderivative by integrating the individual terms after using partial fraction decomposition. Finally, we evaluated the antiderivative using the limits and computed the value of the integral, which is: $$\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}} d x = -\frac{1}{6} \ln \frac{2}{65}$$

Step-by-step solution

Simplify the integrand

Observe that we can factor out \(e^{3x}\) from both the numerator and the denominator: $$\frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}} = \frac{e^{3x}(1 - e^{-6x})}{e^{3x}(1 + e^{-6x})}$$ This allows us to cancel out \(e^{3x}\) from both the numerator and the denominator, resulting in a simpler integrand: $$\int_{0}^{\ln 2} \frac{1 - e^{-6x}}{1 + e^{-6x}} d x$$

Apply the substitution method

Let's use substitution method with \(u = e^{-6x}\), which implies \(\frac{du}{dx} = -6e^{-6x}\) and \(dx = \frac{1}{-6} u^{-1} du\) Now we will also change the limits according to the new variable \(u\): when \(x = 0\), \(u = e^{-6\times0} = 1\) when \(x = \ln2\), \(u = e^{-6\times\ln2} = 2^{-6} = \frac{1}{64}\) So, our integral now becomes: $$ \int_{1}^{\frac{1}{64}} \frac{1 - u}{(1 + u)} \frac{1}{-6} u^{-1} du = -\frac{1}{6} \int_{1}^{\frac{1}{64}} \frac{(1 - u) u^{-1}}{(1 + u)} du$$ Now our integral is in a form that is easy to work with.

Find the antiderivative

Now we need to find the antiderivative of the simplified function: $$-\frac{1}{6} \int_{1}^{\frac{1}{64}} \frac{(1 - u) u^{-1}}{(1 + u)} du = -\frac{1}{6} \int_{1}^{\frac{1}{64}} \frac{1 - u}{u(1 + u)} du$$ Performing partial fraction decomposition, we get: $$\frac{1 - u}{u(1 + u)} = \frac{A}{u} + \frac{B}{1 + u}$$ Solving for \(A\) and \(B\), we find that \(A = 1\) and \(B = -1\). Therefore, we can rewrite the integrand as: $$ \frac{1 - u}{u(1 + u)} = \frac{1}{u} - \frac{1}{1 + u} $$ Now we can integrate the individual terms: $$-\frac{1}{6} \int_{1}^{\frac{1}{64}} \left(\frac{1}{u} - \frac{1}{1 + u}\right) du = -\frac{1}{6} \left[\int_{1}^{\frac{1}{64}} \frac{1}{u} du - \int_{1}^{\frac{1}{64}} \frac{1}{1 + u} du\right]$$ Integrating each term, we have: $$ -\frac{1}{6} \left[\ln |u| - \ln |1 + u|\right]$$

Evaluate the antiderivative using the given limits

Now we will evaluate the antiderivative using the limits \(1\) and \(\frac{1}{64}\): $$\begin{aligned} -\frac{1}{6}\left[\ln\left|\frac{1}{64}\right| - \ln\left|1 + \frac{1}{64}\right| - (\ln|1| - \ln|1+1|)\right] \end{aligned}$$

Compute the value of the integral

Now let's compute the value of the integral: $$\begin{aligned} -\frac{1}{6}\left[\ln\frac{1}{64} - \ln\frac{65}{64} - (\ln1 - \ln2)\right] &= -\frac{1}{6}\left(-\ln 64 - \ln\frac{65}{64} + \ln 2\right) \\ &= -\frac{1}{6}\left((-\ln64+\ln2) - \ln\frac{65}{64}\right) \\ &= -\frac{1}{6}\left( \ln\frac{2}{64} - \ln\frac{65}{64}\right) \\ &= -\frac{1}{6} \ln \frac{2}{65} \end{aligned}$$ So the integral evaluates to: $$\int_{0}^{\ln 2} \frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}} d x = -\frac{1}{6} \ln \frac{2}{65}$$

Key concepts

Integration by Substitution

Integration by substitution is a powerful technique for solving integrals. It can simplify complex integrals into more manageable ones. In this technique, we replace a tricky expression with a new variable, simplifying the integrand.

For instance, in the given integral problem, we approached substitution by choosing a new variable:

• Let \( u = e^{-6x} \). This reduces calculations by removing the exponential in the integrand.

Substitution changes both the integrand and the differential \( dx \) to a new expression in terms of \( du \).

Then, remember to adjust the integration limits to match the new variable. In the problem, when \( x = 0 \), \( u = 1 \) and when \( x = \ln 2 \), \( u = \frac{1}{64} \). With substitution, the integral becomes simpler, ready for further steps like partial fraction decomposition.

Definite Integrals

Definite integrals calculate the area under a curve from a starting point to an endpoint. It's denoted by integral signs with limits of integration at the top and bottom.

In our problem, we calculate the integral from \( x = 0 \) to \( x = \ln 2 \). Definite integrals consist of finding the antiderivative and evaluating it between these limits:

• First, find the indefinite integral (antiderivative) of the function.
• Next, substitute the upper and lower limits into the antiderivative and subtract the results.

The step-by-step solution finishes by evaluating the antiderivative at the given limits, resulting in the expression \(-\frac{1}{6} \ln \frac{2}{65}\). This indicates the calculated area under the original curve.

Partial Fraction Decomposition

Partial fraction decomposition is a major concept in calculus for integrating rational functions, functions expressed as a fraction of polynomials. It breaks down a complex fraction into a sum of simpler fractions.

In the provided integral, the simplification led to the fractional expression \( \frac{1 - u}{u(1 + u)} \). By decomposing into partial fractions, we find constants \( A \) and \( B \) to express it as:

• \( \frac{1 - u}{u(1 + u)} = \frac{A}{u} + \frac{B}{1+u} \)

Solving for these constants (\( A = 1 \) and \( B = -1 \)), the integrand becomes easier to integrate separately.

Once in this form, it's straightforward to integrate each term using basic logarithmic integration rules, preparing the problem for a definite evaluation. Partial fractions significantly simplify the computation, making the integral tractable.