標題:
發問:
∫√tan x dx = ?
最佳解答:
Evaluate ∫√tan x dx . Let u = √tan x x = tan^(– 1) u2 dx = (2u/(u^4 + 1)) du ∴∫√tan x dx = ∫(2u2/(u^4 + 1)) du Since u^4 + 1 = u^4 + 2u2 + 1 – 2u2 = (u2 + 1)2 – (√2 u)2 = (u2 – √2 u + 1)(u2 + √2 u + 1) So we let 2u2/(u^4 + 1) ≡ (Au + B)/(u2 – √2 u + 1) + (Cu + D)/(u2 + √2 u + 1) 2u2 ≡ (Au + B)(u2 + √2 u + 1) + (Cu + D)(u2 – √2 u + 1) 2u2 ≡ Au3 + √2 Au2 + Au + Bu2 + √2 Bu + B + Cu3 – √2 Cu2 + Cu + Du2 – √2 Du + D 2u2 ≡ (A + C)u3 + (√2 A + B – √2 C + D)u2 + (A + √2 B + C – √2 D)u + B + D ╭ │A + C = 0……………………....(1) ∴─┤√2 A + B – √2 C + D = 2………(2) │A + √2 B + C – √2 D = 0………(3) │B + D = 0………………………(4) ╰ Put (1) into (3) and (4) into (2) , √2 A – √2 C = 2 A – C = √2…………(5) √2 B – √2 D = 0 B – D = 0…………...(6) Solving (1) , (4) , (5) and (6) , we get A = (√2)/2 , B = 0 , C = – (√2)/2 and D = 0 ∴∫(2u2/(u^4 + 1)) du = ((√2)/2) ∫(u/(u2 – √2 u + 1)) du – ((√2)/2) ∫(u/(u2 + √2 u + 1)) du = ((√2)/4) ∫(2u/(u2 – √2 u + 1)) du – ((√2)/4) ∫(2u/(u2 + √2 u + 1)) du = ((√2)/4) ∫((2u – √2)/(u2 – √2 u + 1)) du + (1/2) ∫(du/(u2 – √2 u + 1)) – ((√2)/4) ∫((2u + √2)/(u2 + √2 u + 1)) du + (1/2) ∫(du/(u2 + √2 u + 1)) = ((√2)/4) ∫((d(u2 – √2 u + 1))/(u2 – √2 u + 1)) – ((√2)/4) ∫((d(u2 + √2 u + 1))/(u2 + √2 u + 1)) + (1/2) ∫(du/((u + (√2)/2)2 + 1/2)) + (1/2) ∫(du/((u – (√2)/2)2 + 1/2)) Let u + (√2)/2 = ((√2)/2) tan θ , u – (√2)/2 = ((√2)/2) tan φ du = ((√2)/2) sec2 θ dθ , du = ((√2)/2) sec2 φ dφ ∴((√2)/4) ∫((d(u2 – √2 u + 1))/(u2 – √2 u + 1)) – ((√2)/4) ∫((d(u2 + √2 u + 1))/(u2 + √2 u + 1)) + (1/2) ∫(du/((u + (√2)/2)2 + 1/2)) + (1/2) ∫(du/((u – (√2)/2)2 + 1/2)) = ((√2)/4) ln (u2 – √2 u + 1) – ((√2)/4) ln (u2 + √2 u + 1) + (1/2) ∫([((√2)/2) sec2 θ dθ]/[(((√2)/2) tan θ)2 + 1/2]) + (1/2) ∫([((√2)/2) sec2 φ dφ]/[(((√2)/2) tan φ)2 + 1/2]) + C_1 = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (1/2) ∫([((√2)/2) sec2 θ dθ]/[(1/2) sec2 θ]) + (1/2) ∫([((√2)/2) sec2 φ dφ]/[(1/2) sec2 φ]) + C_1 = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (1/2) ∫√2 dθ + (1/2) ∫√2 dφ + C_1 = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (√2 θ)/2 + (√2 φ)/2 + C = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (tan^(– 1) (√2 u + 1) + tan^(– 1) (√2 u – 1))/√2 + C = ((√2)/4) ln ((tan x – √(2 tan x) + 1)/(tan x + √(2 tan x) + 1)) + (tan^(– 1) (√(2 tan x) + 1) + tan^(– 1) (√(2 tan x) – 1))/√2 + C
其他解答:
You really want to find it? The steps are very cubersome. I show you the answer here first. See if you're really interested. ∫√tan x dx = 2 ^ -3/2 { -2 tan^-1[ 1 - (2^1/2) tan^(1/2) (x) ] + 2 tan^-1[ ( 2^(1/2) tan^(-1/2) (x) + 1 ] + ln[ - tan(x) + 2^(1/2) tan^(1/2) (x) - 1 ] - ln[ tan(x) + 2^(1/2) tan^(1/2) (x) + 1 ] } Differentiate this expression with respect to x, we can get √tan x Pardon me if I have typo.
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∫√tan x dx = ?發問:
∫√tan x dx = ?
最佳解答:
Evaluate ∫√tan x dx . Let u = √tan x x = tan^(– 1) u2 dx = (2u/(u^4 + 1)) du ∴∫√tan x dx = ∫(2u2/(u^4 + 1)) du Since u^4 + 1 = u^4 + 2u2 + 1 – 2u2 = (u2 + 1)2 – (√2 u)2 = (u2 – √2 u + 1)(u2 + √2 u + 1) So we let 2u2/(u^4 + 1) ≡ (Au + B)/(u2 – √2 u + 1) + (Cu + D)/(u2 + √2 u + 1) 2u2 ≡ (Au + B)(u2 + √2 u + 1) + (Cu + D)(u2 – √2 u + 1) 2u2 ≡ Au3 + √2 Au2 + Au + Bu2 + √2 Bu + B + Cu3 – √2 Cu2 + Cu + Du2 – √2 Du + D 2u2 ≡ (A + C)u3 + (√2 A + B – √2 C + D)u2 + (A + √2 B + C – √2 D)u + B + D ╭ │A + C = 0……………………....(1) ∴─┤√2 A + B – √2 C + D = 2………(2) │A + √2 B + C – √2 D = 0………(3) │B + D = 0………………………(4) ╰ Put (1) into (3) and (4) into (2) , √2 A – √2 C = 2 A – C = √2…………(5) √2 B – √2 D = 0 B – D = 0…………...(6) Solving (1) , (4) , (5) and (6) , we get A = (√2)/2 , B = 0 , C = – (√2)/2 and D = 0 ∴∫(2u2/(u^4 + 1)) du = ((√2)/2) ∫(u/(u2 – √2 u + 1)) du – ((√2)/2) ∫(u/(u2 + √2 u + 1)) du = ((√2)/4) ∫(2u/(u2 – √2 u + 1)) du – ((√2)/4) ∫(2u/(u2 + √2 u + 1)) du = ((√2)/4) ∫((2u – √2)/(u2 – √2 u + 1)) du + (1/2) ∫(du/(u2 – √2 u + 1)) – ((√2)/4) ∫((2u + √2)/(u2 + √2 u + 1)) du + (1/2) ∫(du/(u2 + √2 u + 1)) = ((√2)/4) ∫((d(u2 – √2 u + 1))/(u2 – √2 u + 1)) – ((√2)/4) ∫((d(u2 + √2 u + 1))/(u2 + √2 u + 1)) + (1/2) ∫(du/((u + (√2)/2)2 + 1/2)) + (1/2) ∫(du/((u – (√2)/2)2 + 1/2)) Let u + (√2)/2 = ((√2)/2) tan θ , u – (√2)/2 = ((√2)/2) tan φ du = ((√2)/2) sec2 θ dθ , du = ((√2)/2) sec2 φ dφ ∴((√2)/4) ∫((d(u2 – √2 u + 1))/(u2 – √2 u + 1)) – ((√2)/4) ∫((d(u2 + √2 u + 1))/(u2 + √2 u + 1)) + (1/2) ∫(du/((u + (√2)/2)2 + 1/2)) + (1/2) ∫(du/((u – (√2)/2)2 + 1/2)) = ((√2)/4) ln (u2 – √2 u + 1) – ((√2)/4) ln (u2 + √2 u + 1) + (1/2) ∫([((√2)/2) sec2 θ dθ]/[(((√2)/2) tan θ)2 + 1/2]) + (1/2) ∫([((√2)/2) sec2 φ dφ]/[(((√2)/2) tan φ)2 + 1/2]) + C_1 = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (1/2) ∫([((√2)/2) sec2 θ dθ]/[(1/2) sec2 θ]) + (1/2) ∫([((√2)/2) sec2 φ dφ]/[(1/2) sec2 φ]) + C_1 = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (1/2) ∫√2 dθ + (1/2) ∫√2 dφ + C_1 = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (√2 θ)/2 + (√2 φ)/2 + C = ((√2)/4) ln ((u2 – √2 u + 1)/(u2 + √2 u + 1)) + (tan^(– 1) (√2 u + 1) + tan^(– 1) (√2 u – 1))/√2 + C = ((√2)/4) ln ((tan x – √(2 tan x) + 1)/(tan x + √(2 tan x) + 1)) + (tan^(– 1) (√(2 tan x) + 1) + tan^(– 1) (√(2 tan x) – 1))/√2 + C
其他解答:
You really want to find it? The steps are very cubersome. I show you the answer here first. See if you're really interested. ∫√tan x dx = 2 ^ -3/2 { -2 tan^-1[ 1 - (2^1/2) tan^(1/2) (x) ] + 2 tan^-1[ ( 2^(1/2) tan^(-1/2) (x) + 1 ] + ln[ - tan(x) + 2^(1/2) tan^(1/2) (x) - 1 ] - ln[ tan(x) + 2^(1/2) tan^(1/2) (x) + 1 ] } Differentiate this expression with respect to x, we can get √tan x Pardon me if I have typo.
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