The given integral to evaluate is:

$\int e^{4x} \sqrt{1 + e^{2x}} \, dx$

Step-by-step solution:

Step 1: Substitution

We are instructed to use trigonometric substitution. Let: $u = e^x \implies du = e^x \, dx$

This means $e^{2x} = u^2$, $e^x \, dx = du$, and the integral becomes: $\int e^{4x} \sqrt{1 + e^{2x}} \, dx = \int u^4 \sqrt{1 + u^2} \, du$

Step 2: Simplify the substitution

Use $v = \sqrt{1 + u^2}$, which implies: $v^2 = 1 + u^2 \quad \text{and} \quad dv = \frac{u \, du}{\sqrt{1 + u^2}} = \frac{u}{v} \, du$

From $v^2 = 1 + u^2$, rearranging gives: $u^2 = v^2 - 1$

Substitute $u^2 = v^2 - 1$ and $du = v \, dv / u$ into the integral: $\int u^4 \sqrt{1 + u^2} \, du = \int (v^2 - 1)^2 v \, dv$

Step 3: Expand and simplify

Expand $(v^2 - 1)^2$: $(v^2 - 1)^2 = v^4 - 2v^2 + 1$

Substitute back into the integral: $\int (v^2 - 1)^2 v \, dv = \int (v^4 - 2v^2 + 1) v \, dv = \int (v^5 - 2v^3 + v) \, dv$

Step 4: Integrate term by term

$\int (v^5 - 2v^3 + v) \, dv = \frac{v^6}{6} - \frac{2v^4}{4} + \frac{v^2}{2} + C = \frac{v^6}{6} - \frac{v^4}{2} + \frac{v^2}{2} + C$

Step 5: Back-substitute for $v$ and $u$

Recall that $v = \sqrt{1 + u^2}$ and $u = e^x$. Substitute back: $v^2 = 1 + u^2 = 1 + e^{2x}, \quad v = \sqrt{1 + e^{2x}}$

Thus: $\int e^{4x} \sqrt{1 + e^{2x}} \, dx = \frac{(1 + e^{2x})^3}{6} - \frac{(1 + e^{2x})^2}{2} + \frac{1 + e^{2x}}{2} + C$

This is the final answer in terms of $x$.

Would you like additional details or have any questions?

Related Questions:

1. How does the substitution $u = e^x$ simplify the integral?
2. Can this problem be solved using hyperbolic substitutions instead?
3. What would the solution look like in terms of definite integrals with given bounds?
4. What is the purpose of using trigonometric substitution here?
5. How does the integral change if a different substitution is applied?

Tip:

When dealing with square roots of quadratic expressions, trigonometric or hyperbolic substitutions often simplify the problem significantly. Always check for these strategies!