The given image represents a definite integral expression:

$-\ln 2 \int e^\theta 2 \cosh(\theta) \, d\theta - \ln 6$

To solve this, let's break it down step-by-step:

1. First, consider the integral part:

$I = \int 2 e^\theta \cosh(\theta) \, d\theta$

Recall that:

$\cosh(\theta) = \frac{e^\theta + e^{-\theta}}{2}$

Substitute this into the integral:

$I = \int 2 e^\theta \left( \frac{e^\theta + e^{-\theta}}{2} \right) \, d\theta$

Simplify:

$I = \int e^\theta (e^\theta + e^{-\theta}) \, d\theta$

Now distribute:

$I = \int e^{2\theta} \, d\theta + \int 1 \, d\theta$

Solve each part:

• The first integral is:

$\int e^{2\theta} \, d\theta = \frac{e^{2\theta}}{2}$

• The second integral is simply:

$\int 1 \, d\theta = \theta$

Thus, the result of the integral is:

$I = \frac{e^{2\theta}}{2} + \theta + C$

2. Next, return to the full expression:

$-\ln 2 \left( \frac{e^{2\theta}}{2} + \theta \right) - \ln 6$

Distribute $-\ln 2$:

$-\ln 2 \cdot \frac{e^{2\theta}}{2} - \ln 2 \cdot \theta - \ln 6$

So, the final result is:

$-\frac{\ln 2}{2} e^{2\theta} - \ln 2 \cdot \theta - \ln 6 + C$

Would you like further details or clarification?

Relative questions:

1. What is the meaning of the hyperbolic cosine function $\cosh(\theta)$ and how is it derived?
2. How do we compute definite integrals when there are logarithmic coefficients?
3. How does the distribution of constants work in integration problems?
4. What are some practical applications of the hyperbolic cosine function in physics or engineering?
5. Why is understanding substitution methods essential for solving integrals?

Tip:

When dealing with integrals of exponential and hyperbolic functions, always try to rewrite them in simpler forms like exponentials to make the integration more straightforward.