   # Deriving a Sum Derive Euler’s famous result that was mentioned in Section 9.3, ∑ n = 1 ∞ 1 n 2 = π 2 6 by completing each step. (a) Prove that ∫ d v 2 − u 2 + v 2 = 1 2 − u 2 arctan v 2 − u 2 + C (b) Prove that I 1 = ∫ 0 2 / 2 ∫ − u u 2 2 − u 2 + v 2 d v d u = π 2 18 by using the substitution u = 2 sin θ . (c) Prove that I 2 = ∫ 2 / 2 2 ∫ u − 2 − u + 2 2 2 − u 2 + v 2 d v d u = 4 ∫ π / 6 π / 2 arctan 1 − sin θ cos θ d θ by using the substitution u = 2 sin θ . (d) Prove the trigonometric identity 1 − sin θ cos θ = tan [ ( π / 2 ) − θ 2 ] . (e) Prove that I 2 = ∫ 2 / 2 2 ∫ u − 2 − u + 2 2 2 − u 2 + v 2 d v d u = π 2 9 . (f) Use the formula for the sum of an infinite geometric series to verify that ∑ n = 1 ∞ 1 n 2 = ∫ 0 1 ∫ 0 1 1 1 − x y d x d y . (g) Use the change of variables u = x + y 2 and v = y − x 2 to prove that ∑ n = 1 ∞ 1 n 2 = ∫ 0 1 ∫ 0 1 1 1 − x y d x d y = I 1 + I 2 = π 2 6 . ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378

#### Solutions

Chapter
Section ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 14, Problem 5PS
Textbook Problem
1 views

## Deriving a Sum Derive Euler’s famous result that was mentioned in Section 9.3, ∑ n = 1 ∞ 1 n 2 = π 2 6 by completing each step.(a) Prove that ∫ d v 2 − u 2 + v 2 = 1 2 − u 2 arctan v 2 − u 2 + C (b) Prove that I 1 = ∫ 0 2 / 2 ∫ − u u 2 2 − u 2 + v 2 d v   d u = π 2 18 by using the substitution u = 2 sin θ .(c) Prove that I 2 = ∫ 2 / 2 2 ∫ u − 2 − u + 2 2 2 − u 2 + v 2 d v   d u       = 4 ∫ π / 6 π / 2 arctan 1 − sin θ cos θ d θ by using the substitution u = 2 sin θ .(d) Prove the trigonometric identity 1 − sin θ cos θ = tan [ ( π / 2 ) − θ 2 ] .(e) Prove that I 2 = ∫ 2 / 2 2 ∫ u − 2 − u + 2 2 2 − u 2 + v 2 d v   d u = π 2 9 .(f) Use the formula for the sum of an infinite geometric series to verify that ∑ n = 1 ∞ 1 n 2 = ∫ 0 1 ∫ 0 1 1 1 − x y   d x   d y .(g) Use the change of variables u = x + y 2 and v = y − x 2 to prove that ∑ n = 1 ∞ 1 n 2 = ∫ 0 1 ∫ 0 1 1 1 − x y d x d y = I 1 + I 2 = π 2 6 .

(a)

To determine

To prove: The given equation dv2u2+v2=12u2arctan(v2u2)+c.

### Explanation of Solution

Given:

The equation provided is,

dv2u2+v2=12u2arctan(v2u2)+c

Formula used:

Integration formula,

dxa2+x2=1aarctan(xa)+c

Proof:

Take the provided equation into consideration.

dv2u2+v2

The above equation can be rewritten as,



(b)

To determine

To prove: The given equation I1=022uu22u2v2dvdu=π218 by substituting u=2sinθ.

(c)

To determine

To prove: The provided equation I2=222u2u+222u2+v2dvdu=4π6π2arctan(1sinθcosθ)dθ by substituting u=2sinθ.

(d)

To determine

To prove: The provided trigonometric equation 1sinθcosθ=tan(π2θ2).

(e)

To determine

To prove: The provided equation I2=222u2u+222u2+v2dvdu=π29.

(f)

To determine

To prove: The provided equation n=11n2=010111xydxdy.

(g)

To determine

To prove: The expression:

n=11n2=010111xydxdy=I1+I2=π26, using change of variables, u=x+y2,v=xy2.

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