(a) Prove that dv 1 2 u? + v2 arctan u- + C. 2 u 2, /2/2 ru (b) Prove that I dv du 2 - u? + v2 18 by using the substitution u = 2 sin 0. (c) Prove that 2 I, = dv du %3D u? + v2 2/2 Ju-2 1/2 1 sin 0 de cos e - = 4 arctan Jn/6 by using the substitution u = /2 sin 0.
(a) Prove that dv 1 2 u? + v2 arctan u- + C. 2 u 2, /2/2 ru (b) Prove that I dv du 2 - u? + v2 18 by using the substitution u = 2 sin 0. (c) Prove that 2 I, = dv du %3D u? + v2 2/2 Ju-2 1/2 1 sin 0 de cos e - = 4 arctan Jn/6 by using the substitution u = /2 sin 0.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I need help figuring out parts a-c for this problem. Thanks!
![### 5. Deriving a Sum
**Objective**: Derive Euler’s famous result that was mentioned in Section 9.3:
\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
\]
by completing each step.
---
**(a)** Prove that
\[
\int \frac{dv}{2 - u^2 + v^2} = \frac{1}{\sqrt{2 - u^2}} \arctan \frac{v}{\sqrt{2 - u^2}} + C.
\]
---
**(b)** Prove that
\[
I_1 = \int_0^{\sqrt{2}/2} \int_{-u}^{u} \frac{2}{2 - u^2 + v^2} \, dv \, du = \frac{\pi^2}{18}
\]
by using the substitution \( u = \sqrt{2} \sin \theta \).
---
**(c)** Prove that
\[
I_2 = \int_{\sqrt{2}/2}^{\sqrt{2}} \int_{-\sqrt{2}}^{-u+\sqrt{2}} \frac{2}{2 - u^2 + v^2} \, dv \, du
\]
\[
= 4 \int_{\pi/6}^{\pi/2} \arctan \frac{1 - \sin \theta}{\cos \theta} \, d\theta
\]
by using the substitution \( u = \sqrt{2} \sin \theta \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fda21d2aa-bb54-4a64-8baf-73046aeb648e%2Fc9d4a583-2fd6-4ac2-8f49-27cbf68372c0%2Fdh9ey1m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### 5. Deriving a Sum
**Objective**: Derive Euler’s famous result that was mentioned in Section 9.3:
\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
\]
by completing each step.
---
**(a)** Prove that
\[
\int \frac{dv}{2 - u^2 + v^2} = \frac{1}{\sqrt{2 - u^2}} \arctan \frac{v}{\sqrt{2 - u^2}} + C.
\]
---
**(b)** Prove that
\[
I_1 = \int_0^{\sqrt{2}/2} \int_{-u}^{u} \frac{2}{2 - u^2 + v^2} \, dv \, du = \frac{\pi^2}{18}
\]
by using the substitution \( u = \sqrt{2} \sin \theta \).
---
**(c)** Prove that
\[
I_2 = \int_{\sqrt{2}/2}^{\sqrt{2}} \int_{-\sqrt{2}}^{-u+\sqrt{2}} \frac{2}{2 - u^2 + v^2} \, dv \, du
\]
\[
= 4 \int_{\pi/6}^{\pi/2} \arctan \frac{1 - \sin \theta}{\cos \theta} \, d\theta
\]
by using the substitution \( u = \sqrt{2} \sin \theta \).
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