To solve the given integral using $u$-substitution, follow these steps:

The integral is:

$\int 2^x \left[1 - 3^x \cdot \text{csch}(6^x)\right] dx$

1. Substitution for simplification: Let $u = 2^x$. Then, the derivative is: $du = 2^x \ln(2) dx \quad \text{or} \quad dx = \frac{du}{u \ln(2)}.$

2. Substitute $2^x = u$ and adjust $dx$ in the integral: [ \int 2^x \left[1 - 3^x \cdot \text{csch}(6^x)\right] dx = \int u \left[1 - 3^x \cdot \text{csch}(6^x)\right] \frac{du}{u \ln(2)}. ]

   Simplify: $\int \frac{u}{\ln(2)} \left[1 - 3^x \cdot \text{csch}(6^x)\right] \frac{du}{u}.$

3. Further substitution: Recognize $3^x$ can be handled with a separate substitution: Let $v = 3^x$, so $dv = 3^x \ln(3) dx$.

   Similarly, $6^x = (2^x \cdot 3^x)$, and this links $u$ and $v$ within the $\text{csch}(6^x)$.

From here, the integration becomes more manageable step by step with the $u$- and $v$-substitutions depending on the part being analyzed.

Would you like detailed substitutions carried through further or clarifications on specific steps?

Related Questions:

1. What is the role of logarithmic differentiation in $u$-substitution?
2. How do hyperbolic trigonometric functions like $\text{csch}(x)$ transform under substitution?
3. Can substitution methods always simplify integrals with exponential terms?
4. How does $u = 2^x$ help link different exponential bases in an integral?
5. What is the importance of recognizing the derivative relationships for substitution?

Tip:

Always ensure that the substitution covers the entire integral (including limits if it's definite) to avoid leaving terms undefined.