Give step-by-step solution with explanation and final answer:)Evaluate[(9 -e #2x) ~1/2
Question:
Give step-by-step solution with explanation and final answer:
)Evaluate[(9 -e #2x) ~1/2
This is AI generated Response.
Asked by: gvr
Created at: 2025-06-13 03:45:37
Asked by: gvr
Created at: 2025-06-13 03:45:37
Final Answer:
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Let's evaluate the integral:
\[
\int \sqrt{9 - e^{2x}} \, dx
\]
### Step 1: **Substitution**
Let \( u = e^x \implies du = e^x dx \implies dx = \frac{du}{u} \).
Also, \( e^{2x} = (e^x)^2 = u^2 \).
Now substitute into the integral:
\[
\int \sqrt{9 - e^{2x}} \, dx = \int \sqrt{9 - u^2} \cdot \frac{du}{u}
\]
### Step 2: **Rewrite the Integral**
\[
\int \frac{\sqrt{9 - u^2}}{u} \, du
\]
Let's use the substitution \( u = 3\sin\theta \):
- \( du = 3\cos\theta\, d\theta \)
- \( 9 - u^2 = 9 - 9\sin^2\theta = 9\cos^2\theta \)
- \( \sqrt{9 - u^2} = 3\cos\theta \)
Now substitute:
\[
\int \frac{3\cos\theta}{3\sin\theta} \cdot 3\cos\theta\, d\theta
= \int \frac{3\cos\theta \cdot 3\cos\theta}{3\sin\theta} d\theta
= 3 \int \frac{\cos^2\theta}{\sin\theta} d\theta
\]
### Step 3: **Simplify the Integral**
\[
3 \int \frac{\cos^2\theta}{\sin\theta} d\theta
\]
Recall \( \cos^2\theta = 1 - \sin^2\theta \):
\[
= 3 \int \frac{1 - \sin^2\theta}{\sin\theta} d\theta
= 3 \int \left( \frac{1}{\sin\theta} - \sin\theta \right) d\theta
= 3 \int \csc\theta\, d\theta - 3\int \sin\theta\, d\theta
\]
### Step 4: **Integrate**
- \( \int \csc\theta\, d\theta = \ln|\tan\frac{\theta}{2}| \)
- \( \int \sin\theta\, d\theta = -\cos\theta \)
Plug these in:
\[
= 3 \ln|\tan\frac{\theta}{2}| + 3\cos\theta + C
\]
### Step 5: **Back Substitute**
Recall \( u = 3\sin\theta \implies \sin\theta = \frac{u}{3} \implies \theta = \arcsin\left(\frac{u}{3}\right) \).
Also, \( u = e^x \):
So,
- \( \theta = \arcsin\left(\frac{e^x}{3}\right) \)
- \( \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{e^x}{3}\right)^2} = \frac{\sqrt{9 - e^{2x}}}{3} \)
So,
\[
\boxed{
\int \sqrt{9 - e^{2x}}\, dx = 3 \ln\left|\tan\left(\frac{1}{2}\arcsin\left(\frac{e^x}{3}\right)\right)\right| + \sqrt{9 - e^{2x}} + C
}
\]
---
**Final Answer:**
\[
\boxed{
\int \sqrt{9 - e^{2x}}\, dx = 3 \ln\left|\tan\left(\frac{1}{2}\arcsin\left(\frac{e^x}{3}\right)\right)\right| + \sqrt{9 - e^{2x}} + C
}
\]
Question:
i need 2 lines pure theory for explanation
This is AI generated Response.
Asked by: gvr
Created at: 2025-06-13 03:46:24
Asked by: gvr
Created at: 2025-06-13 03:46:24
Final Answer:
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The integral involves a substitution to simplify the square root expression, transforming it into a trigonometric form that allows for standard integration techniques. By expressing the variable in terms of sine and cosine functions, the integral becomes manageable and leads to the final solution involving inverse sine and tangent functions.
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