8.3.43) Queetion: Tind the volumeof the solid. The region bounded ou f-e-* X=In2, and the coordinate axes brevolved about trev |(tan 4a)(sec 4x – 1) da = | (tan 4r) sec² 4x da – tan 4r dr = tan 4r dr axis.
8.3.43) Queetion: Tind the volumeof the solid. The region bounded ou f-e-* X=In2, and the coordinate axes brevolved about trev |(tan 4a)(sec 4x – 1) da = | (tan 4r) sec² 4x da – tan 4r dr = tan 4r dr axis.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The question is circled in bright yellow. My question is, where did the tan^3 4xdx come from?
![### Section 8.3.43
**Question:** Find the volume of the solid. The region bounded by \( f(x) = e^{-x} \), \( x = \ln 2 \), and the coordinate axes is revolved about the y-axis.
#### Solution:
To solve the given problem, we first address the integration involved:
\[
\int \tan^3 4x \, dx = \int (\tan 4x)(\sec^2 4x - 1) \, dx = \int (\tan 4x) \sec^2 4x \, dx - \int \tan 4x \, dx
\]
This can be rewritten as:
\[
= \int (\tan 4x) \sec^2 4x \, dx + \frac{\ln |\cos 4x|}{4} + C
\]
Now, let's substitute:
Let \( u = \tan 4x \) so that \( du = 4 \sec^2 4x \, dx \). Substituting gives:
\[
\frac{1}{4} \int u \, du + \frac{\ln |\cos 4x|}{4} + C = \frac{u^2}{8} + \frac{\ln |\cos 4x|}{4} + C
\]
This can be further simplified to:
\[
= \frac{\tan^2 4x}{8} + \frac{\ln |\cos 4x|}{4} + C
\]
This step-by-step solution shows the integration process used to find the area under the curve, which is necessary to compute the volume of the solid revolved about the y-axis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F0554994c-4f73-424c-ac78-7d312d7265cb%2Fu5luwa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Section 8.3.43
**Question:** Find the volume of the solid. The region bounded by \( f(x) = e^{-x} \), \( x = \ln 2 \), and the coordinate axes is revolved about the y-axis.
#### Solution:
To solve the given problem, we first address the integration involved:
\[
\int \tan^3 4x \, dx = \int (\tan 4x)(\sec^2 4x - 1) \, dx = \int (\tan 4x) \sec^2 4x \, dx - \int \tan 4x \, dx
\]
This can be rewritten as:
\[
= \int (\tan 4x) \sec^2 4x \, dx + \frac{\ln |\cos 4x|}{4} + C
\]
Now, let's substitute:
Let \( u = \tan 4x \) so that \( du = 4 \sec^2 4x \, dx \). Substituting gives:
\[
\frac{1}{4} \int u \, du + \frac{\ln |\cos 4x|}{4} + C = \frac{u^2}{8} + \frac{\ln |\cos 4x|}{4} + C
\]
This can be further simplified to:
\[
= \frac{\tan^2 4x}{8} + \frac{\ln |\cos 4x|}{4} + C
\]
This step-by-step solution shows the integration process used to find the area under the curve, which is necessary to compute the volume of the solid revolved about the y-axis.
Expert Solution
Step 1
No need for this term for evaluating the volume of the revolved solid.
Step by step
Solved in 2 steps with 1 images
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