Integrate using Trig Sub Integral of dx/(4+x^2)?
回答 (2)
2018-01-08 11:47 pm
✔ 最佳答案
1 + tan(t)^2 = sec(t)^24 + x^2 = 4 + 4 * tan(t)^2
x^2 = 4 * tan(t)^2
x = 2 * tan(t)
dx = 2 * sec(t)^2 * dt
dx / (4 + x^2) =>
2 * sec(t)^2 * dt / (4 + 4 * tan(t)^2) =>
2 * sec(t)^2 * dt / (4 * sec(t)^2) =>
dt / 2
Integrate
(1/2) * t + C
x = 2 * tan(t)
x/2 = tan(t)
t = arctan(x/2)
(1/2) * arctan(x/2) + C
2018-01-09 12:08 am
Put x = 2 tan(u)
Then, dx = 2 sec²(u) du
and u = arctan(x/2)
∫ dx / (4 + x²)
= ∫ 2 sec²(u) du / {4 + [2 tan(u)]²}
= ∫ 2 sec²(u) du / [4 + 4 tan²(u)]
= ∫ 2 sec²(u) du / {4 [1 + tan²(u)]}
= (1/2) ∫ sec²(u) du / [1 + tan²(u)]
= (1/2) ∫ sec²(u) du / sec²(u)
= (1/2) ∫ du
= (1/2) u + C
= (1/2) arctan(x/2) + C
Then, dx = 2 sec²(u) du
and u = arctan(x/2)
∫ dx / (4 + x²)
= ∫ 2 sec²(u) du / {4 + [2 tan(u)]²}
= ∫ 2 sec²(u) du / [4 + 4 tan²(u)]
= ∫ 2 sec²(u) du / {4 [1 + tan²(u)]}
= (1/2) ∫ sec²(u) du / [1 + tan²(u)]
= (1/2) ∫ sec²(u) du / sec²(u)
= (1/2) ∫ du
= (1/2) u + C
= (1/2) arctan(x/2) + C
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