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#1
Report Thread starter 14 years ago
#1
Integrate:

a) 3e^(2x-3) dx

b) 1/(x^2+x^4) dx

c) 2/(9-x^2)^1/2 dx

d) (2cosx+1)(2cosx-1) dx


Thanks in advance.
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Jonny W
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#2
Report 14 years ago
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(a)
If you can integrate e^(2x) in your head then you can do this one easily by writing e^(2x - 3) as e^(2x)*e(-3).

u = 2x - 3
du/dx = 2
(int) 3e^(2x - 3) dx
= (int) (3/2)e^u du
= (3/2)e^u + c
= (3/2)e^(2x - 3) + c

(b)
1/(x^2 + x^4)
= 1/(x^2(1 + x^2))
= 1/x^2 - 1/(1 + x^2)

(int) 1/(x^2 + x^4) dx = -1/x - arctan(x) + c

(c)
x = 3sin(t)
dx/dt = 3cos(t)

(int) 2 / sqrt(9 - x^2) dx
= (int) 6cos(t) / (3cos(t)) dt
= (int) 2 dt
= 2t + c
= 2arcsin(x/3) + c

(d)
(2cos(x) + 1)(2cos(x) - 1)
= 4cos^2(x) - 1
= 2(cos(2x) + 1) - 1
= 2cos(2x) + 1

(int) (2cos(x) + 1)(2cos(x) - 1) dx = sin(2x) + x + c
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J.F.N
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#3
Report 14 years ago
#3
a) 3/2(e^2x-3) + F
b) Split as partial fractions to get 1/x^2 - 1/(1-x^2). These can be integrated to give -1/x - arctanx + U
c) 2arcsin(x/3) + C
d) I didn't get this instantly, and I don't have time to think about it. I'm sure someone else will help.
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