6:47 PM Mon 17 Mar 98\% mylab.pearson.com Kurulus Osm... Can't Open P... CUNY Login The Evolutio... Discord Do Homewor... MAT 220-S25 Rasheed Larry 03/17/25 6:47 PM Homework: Trigonometric Question 4, 8.4.35 HW Score: 33.33\%, 3 Substitutions of 9 points ○ Points: 0 of 1 Save Question list Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Use an appropriate substitution and then a trigonometric substitution to evaluate the integral. $$ \int_{0}^{\ln 3} \frac{e^{\mathrm{t}} \mathrm{dt}}{\sqrt{e^{2 \mathrm{t}}+16}} $$ $$ \int_{0}^{\ln 3} \frac{e^{t} d t}{\sqrt{e^{2 t}+16}}=\ln \left(\frac{\sqrt{17-1}}{2}\right) $$ (Type an exact answer, using radicals as needed.)

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To evaluate the integral

$$
\int_{0}^{\ln 3} \frac{e^{t} dt}{\sqrt{e^{2 t}+16}},
$$

we will use a substitution followed by a trigonometric substitution. Let's break this down step by step.

### Step 1: Substitution

Let $$ u = e^t $$. Then, the differential $$ du = e^t dt $$ or $$ dt = \frac{du}{u} $$. The limits of integration change as follows:
- When $$ t = 0 $$, $$ u = e^0 = 1 $$.
- When $$ t = \ln 3 $$, $$ u = e^{\ln 3} = 3 $$.

Now, we can rewrite the integral in terms of $$ u $$:

$$
\int_{1}^{3} \frac{u \cdot \frac{du}{u}}{\sqrt{u^2 + 16}} = \int_{1}^{3} \frac{du}{\sqrt{u^2 + 16}}.
$$

### Step 2: Trigonometric Substitution

Next, we will use the trigonometric substitution $$ u = 4\tan(\theta) $$. Then, $$ du = 4\sec^2(\theta) d\theta $$ and the expression under the square root becomes:

$$
\sqrt{u^2 + 16} = \sqrt{(4\tan(\theta))^2 + 16} = \sqrt{16\tan^2(\theta) + 16} = \sqrt{16(\tan^2(\theta) + 1)} = 4\sec(\theta).
$$

Now, substituting these into the integral gives:

$$
\int \frac{4\sec^2(\theta) d\theta}{4\sec(\theta)} = \int \sec(\theta) d\theta.
$$

### Step 3: Evaluate the Integral

The integral of $$ \sec(\theta) $$ is:

$$
\int \sec(\theta) d\theta = \ln | \sec(\theta) + \tan(\theta) | + C.
$$

### Step 4: Change Back to $$ u $$

Now we need to change back to $$ u $$. Recall that:

$$
\tan(\theta) = \frac{u}{4} \quad \text{and} \quad \sec(\theta) = \frac{\sqrt{u^2 + 16}}{4}.
$$

Thus,

$$
\sec(\theta) + \tan(\theta) = \frac{\sqrt{u^2 + 16}}{4} + \frac{u}{4} = \frac{\sqrt{u^2 + 16} + u}{4}.
$$

### Step 5: Substitute Back into the Integral

Now we can write the integral as:

$$
\int_{1}^{3} \frac{du}{\sqrt{u^2 + 16}} = \left[ \ln \left| \frac{\sqrt{u^2 + 16} + u}{4} \right| \right]_{1}^{3}.
$$

### Step 6: Evaluate the Limits

Now we evaluate at the limits:

1. For $$ u = 3 $$:

$$
\sqrt{3^2 + 16} + 3 = \sqrt{9 + 16} + 3 = \sqrt{25} + 3 = 5 + 3 = 8.
$$

2. For $$ u = 1 $$:

$$
\sqrt{1^2 + 16} + 1 = \sqrt{1 + 16} + 1 = \sqrt{17} + 1.
$$

Now substituting these back into the logarithm:

$$
\ln \left( \frac{8}{4} \right) - \ln \left( \frac{\sqrt{17} + 1}{4} \right) = \ln(2) - \ln \left( \frac{\sqrt{17} + 1}{4} \right).
$$

Using properties of logarithms:

$$
= \ln(2) + \ln(4) - \ln(\sqrt{17} + 1) = \ln \left( \frac{8}{\sqrt{17} + 1} \right).
$$

### Final Answer

Thus, the final answer for the integral is:

$$
\ln \left( \frac{8}{\sqrt{17} + 1} \right).
$$
$$ \ln \left( \frac{8}{\sqrt{17} + 1} \right) $$

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