Certainly! I'll walk you through two challenging problems that involve substitution for exponential, logarithmic, and trigonometric functions.

Problem 1: Exponential/Logarithmic Function Substitution

Question:
Evaluate the integral:
$\int \frac{e^{2x}}{e^{2x} + 1} \, dx$

Solution:

We can simplify this integral using substitution. Let's define:

$u = e^{2x} + 1$

Now, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = 2e^{2x}$

or equivalently,

$du = 2e^{2x} \, dx$

This substitution turns the integral into:

$\int \frac{e^{2x}}{e^{2x} + 1} \, dx = \frac{1}{2} \int \frac{du}{u}$

The integral of $\frac{1}{u}$ is $\ln|u|$. Substituting back $u = e^{2x} + 1$, we get:

$\frac{1}{2} \ln|e^{2x} + 1| + C$

Thus, the final answer is:

$\frac{1}{2} \ln(e^{2x} + 1) + C$

---

Problem 2: Trigonometric Function Substitution

Question:
Evaluate the integral:
$\int \frac{\sin x}{\cos^2 x} \, dx$

Solution:

Here, we can use substitution. Notice that we have $\sin x$ and $\cos^2 x$, so let’s define:

$u = \cos x$

Now, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = -\sin x$

Thus, $du = -\sin x \, dx$. The integral becomes:

$\int \frac{\sin x}{\cos^2 x} \, dx = - \int \frac{du}{u^2}$

We can integrate $\frac{1}{u^2}$ as $-\frac{1}{u}$, so the integral becomes:

$-\left( -\frac{1}{u} \right) + C = \frac{1}{u} + C$

Substitute back $u = \cos x$:

$\frac{1}{\cos x} + C = \sec x + C$

Thus, the final answer is:

$\sec x + C$

---

Summary

1. Exponential/Logarithmic Substitution: Involves recognizing the structure of the function and using substitution like $u = e^{2x} + 1$ to simplify the integral.
2. Trigonometric Substitution: Involves using simple substitutions like $u = \cos x$ to simplify the integral involving trigonometric functions.

---

Would you like further clarification or more examples on either of these? Here are 5 related questions to explore:

1. How do you handle integrals involving $e^x$ and $\ln x$ simultaneously?
2. How can we use substitution in integrals with trigonometric identities?
3. What if the substitution involves a more complex trigonometric function, like $\sin^2 x$?
4. Can you solve integrals that involve higher powers of trigonometric functions using substitution?
5. How would you approach integrals with a combination of exponential and trigonometric functions?

Tip: When dealing with integrals involving exponential, logarithmic, or trigonometric functions, try to spot patterns where substitution can simplify the expression, such as recognizing derivatives of the components within the integral!