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場在 發問於 科學及數學數學 · 5 年前

5question about Definite integral?

evaluate∫_0^(π/4)▒(sec^2⁡ x)/√(1-tan⁡〖^2 x〗 ) dx by substitution tan x=sin y

for a given positive integer n, let In = ∫_0^(π/4)▒1/cos^n⁡x dx express I3 in terms of I1

a)show that ∫_0^π▒x f(sin⁡〖x)dx= π/2 ∫_0^π▒〖f(sin⁡〖x)dx〗 〗〗…(+) hence or otherwise, evaluate ∫_0^π▒(x sin⁡x)/(1+cos^2⁡x ) b)if f is even function, establish a formula corresponding to (+), with sin x replaced by cos x

let f(x)=sin(πx)/(x+1) show that ∫_a^b▒f(x)dx=-∫_(a+1)^(b+1)▒sin(πx)/x dx where a,b≥0

prove that ∫_0^a▒x^3/(a^2+x^2)^5/2 dx=1/a∫_0^(π/4)▒sin^3θ dθ=(8-5√2)/(12a)

更新:

i have ans but i don't know how to do

ans

1.pi/2

2.i3=1/2i1+√2/2

3.pi^2/4

1 個解答

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  • 匿名
    5 年前
    最愛解答

     

    Question:

    1. Evaluate ∫(0,π/4) (sec²x)/√(1 - tan⁡²x) dx by substitution tan x = sin y

    2. For a given positive integer n, let I𝑛 = ∫(0,π/4) 1/cosⁿ⁡x dx, express I₃ in terms of I₁

    3a. Show that ∫(0,π) x f(sin x) dx = π/2 ∫(0,π) f(sin x) dx …... (+)

      hence or otherwise, evaluate ∫(0,π) (x sin ⁡x)/(1 + cos²⁡x) dx

    3b. If f is even function, establish a formula corresponding to (+), with sin x replaced by cos x

    4. let f(x) = sin(πx)/(x + 1), show that ∫(a,b) f(x) dx = - ∫(a + 1,b + 1) sin(πx)/x dx where a, b ≥ 0

    5. Prove that ∫(0,a) x³/(a² + x²)^(5/2) dx = 1/a ∫(0,π/4) sin³θ dθ = (8 - 5√2)/(12a)

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    Solution:

    1.

    Let tan x = sin y, sec²x dx = cos y dy

    When x = 0, y = 0.

    When x = π/4, y = π/2.

    ∫(0,π/4) (sec²x)/√(1 - tan⁡²x) dx

    = ∫(0,π/2) 1/√(1 - sin⁡²y) cos y dy

    = ∫(0,π/2) (cos y)/√(cos⁡²y) dy

    = ∫(0,π/2) (cos y)/(cos⁡ y) dy

    = ∫(0,π/2) dy

    = π/2

    2.

    I𝑛

    = ∫ 1/cosⁿ⁡x dx, from 0 to π/4

    = ∫ (sec²x)/cosⁿ⁻²⁡x dx, from 0 to π/4

    = ∫ 1/cosⁿ⁻²⁡x d(tan x), , from 0 to π/4

    = { [ (tan x)/(cosⁿ⁻²⁡x) ], from 0 to π/4 } - ∫(0,π/4) tan x d(1/cosⁿ⁻²⁡x)

    = { [tan(π/4)]/[cosⁿ⁻²⁡(π/4)] - 0 ] - ∫(0,π/4) (tan x) (- n + 2)/(cosⁿ⁻²⁺¹x) d(cos x)

    = (√2)ⁿ⁻²⁡ + (n - 2) ∫(0,π/4) (tan x)(- sin x)/(cosⁿ⁻¹x) dx

    = (√2)ⁿ⁻² - (n - 2) ∫(0,π/4) (1 - cos²x)/(cosⁿx) dx

    = (√2)ⁿ⁻² - (n - 2) ∫(0,π/4) 1/(cosⁿx) dx + (n - 2) ∫(0,π/4) (cos²x)/(cosⁿx) dx

    = (√2)ⁿ⁻² - (n - 2) ∫(0,π/4) 1/(cosⁿx) dx + (n - 2) ∫(0,π/4) 1/(cosⁿ⁻²x) dx

    ∴ ∫(0,π/4) 1/cosⁿ⁡x dx = (√2)ⁿ⁻² - (n - 2) ∫(0,π/4) 1/(cosⁿx) dx + (n - 2) ∫(0,π/4) 1/(cosⁿ⁻²x) dx

    (1 + n - 2) ∫(0,π/4) 1/cosⁿ⁡x dx = (√2)ⁿ⁻² + (n - 2) I𝑛₋₂

    I𝑛 = (√2)ⁿ⁻²/(n - 1) + [(n - 2)/(n - 1)] I𝑛₋₂

    I₃

    = (√2)³⁻²/(3 - 1) + [(3 - 2)/(3 - 1)] I₃₋₂

    = (√2)/2 + I₁/2

    3a.

    Let x = π - u, dx = -du

    When x = 0, u = π.

    When x = π, u = 0.

    ∫(0,π) x f(sin x) dx

    = ∫(π,0) (π - u) f(sin (π - u)) (-1) du

    = - ∫(π,0) (π - u) f(sin u) du

    = ∫(0,π) (π - u) f(sin u) du

    = π ∫(0,π) f(sin u) du - ∫(0,π) u f(sin u) du

    = π ∫(0,π) f(sin x) dx - ∫(0,π) x f(sin x) dx

    ∴ ∫(0,π) x f(sin x) dx = π ∫(0,π) f(sin x) dx - ∫(0,π) x f(sin x) dx

    (1 + 1) ∫(0,π) x f(sin x) dx = π ∫(0,π) f(sin x) dx

    ∫(0,π) x f(sin x) dx = π/2 ∫(0,π) f(sin x) dx

    ∫(0,π) (x sin ⁡x)/(1 + cos²⁡x) dx

    = ∫(0,π) (x sin ⁡x)/[1 + (1 - sin²⁡x)] dx

    = ∫(0,π) (x sin ⁡x)/(2 - sin²⁡x) dx

    Let f(x) = x/(2 - x²), f(sin x) = (sin x)/(2 - sin²x)

    = ∫(0,π) x f(sin x) dx

    = π/2 ∫(0,π) f(sin x) dx

    = π/2 ∫(0,π) (sin x)/(2 - sin²x) dx

    = π/2 ∫(0,π) (sin x)/(1 + cos²x) dx

    = -π/2 ∫(0,π) 1/(1 + cos²x) d(cos x)

    Let cos x = tan y, d(cos x) = sec²y dy

    When x = 0, tan y = π/4

    When x = π, tan y = -π/4

    = -π/2 ∫(π/4,-π/4) (sec²y)/(1 + tan²y) dy

    = π/2 ∫(-π/4,π/4) dy

    = π/2 (π/4 + π/4)

    = π²/4

    3b.

    If f is even function, f(-x) = f(x)

    Let x = π - u, dx = -du

    When x = 0, u = π.

    When x = π, u = 0.

    ∫(0,π) x f(cos x) dx

    = ∫(π,0) (π - u) f(cos (π - u)) (-1) du

    = ∫(0,π) (π - u) f(- cos u) du

    = π ∫(0,π) f(cos u) du - ∫(0,π) u f(cos u) du

    = π ∫(0,π) f(cos x) dx - ∫(0,π) x f(cos x) dx

    ∴ ∫(0,π) x f(cos x) dx = π ∫(0,π) f(cos x) dx - ∫(0,π) x f(cos x) dx

    (1 + 1) ∫(0,π) x f(cos x) dx = π ∫(0,π) f(cos x) dx

    ∫(0,π) x f(cos x) dx = π/2 ∫(0,π) f(cos x) dx

    4.

    f(x - 1) = sin[π(x - 1)]/(x - 1 + 1) = sin(πx - π)/x = - sin(πx)/x

    Let x = u - 1, dx = du

    When x = a, u = a + 1.

    When x = b, u = b + 1.

    ∫(a,b) f(x) dx

    = ∫(a + 1,b + 1) f(u - 1) du

    = - ∫(a + 1,b + 1) sin(πu)/u du

    = - ∫(a + 1,b + 1) sin(πx)/x dx

    5.

    Prove that ∫(0,a) x³/(a² + x²)^(5/2) dx = 1/a ∫(0,π/4) sin³θ dθ = (8 - 5√2)/(12a)

    Let x = a tan u, dx = a sec²θ dθ

    When x = 0, θ = 0.

    When x = a, θ = π/4.

    ∫(0,a) x³/(a² + x²)^(5/2) dx

    = ∫(0,π/4) (a³ tan³θ)(a sec²θ)/(a² + a² tan²θ)^(5/2) dθ

    = ∫(0,π/4) (a⁴ tan³θ sec²θ)/(a² sec²θ)^(5/2) dθ

    = ∫(0,π/4) (a⁴ tan³θ sec²θ)/(a⁵ sec⁵θ) dθ

    = 1/a ∫(0,π/4) (tan³θ)/(sec³θ) dθ

    = 1/a ∫(0,π/4) sin³θ dθ

    = 1/a ∫(0,π/4) sinθ (1 - cos²θ) dθ

    = -1/a ∫(0,π/4) (1 - cos²θ) d(cos θ)

    = -1/a [ cosθ - (cos³θ)/3 ], from 0 to π/4

    = -1/a [ 1/√2 - 1/(6√2) - 1 + 1/3 ]

    = -1/a [ 5/(6√2) - 2/3 ]

    = -1/a [ (5√2 - 8)/12 ]

    = (8 - 5√2)/(12a)

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