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Integrating e^ax* sin(bx) by parts twice yields a result of the form:

e^ax* (A sin(bx) + B cos(bx))+C

a. Find the constants A and B in terms of a and b. [Hint: don't actually perform the integration by parts]
b. Evaluate the integral of e^ax * cos(bx) by modifying the result in part (a). [Again, it is not necessary to perform integration by parts, as a result is of the same form as that in part(a).]

2007-01-17 09:40:48 · 4 answers · asked by ? 1 in Science & Mathematics Mathematics

can you do this without integration??

2007-01-17 12:16:59 · update #1

4 answers

∫e^ax* sin(bx)dx
let u = sin(bx)dx, du = bcos(bx)dx
dv = e^axdx, v = (1/a)e^ax
∫e^ax*sin(bx)dx = (1/a)(e^ax)sin(bx) - (b/a)∫(e^ax)cos(bx)dx
let u = cos(bx), du = -bsin(bx)dx
dv = (e^ax)dx, v = (1/a)(e^ax)
uv = (1/a)(e^ax)cos(bx),
A = (1/(2a))
B = -(b/(2a^2))

for ∫(e^ax)cos(bx)dx
A = -(1/(2a))
B = (b/(2a^2))

2007-01-17 11:37:17 · answer #1 · answered by Helmut 7 · 0 0

Taking the derivative of the integral should get us back where we started from. So, we'll take the derivative of
e^ax* (A sin(bx) + B cos(bx))+C (the antiderivative)

(e^ax* (A sin(bx) + B cos(bx))+C)'
=(e^ax)'(A sin(bx) + B cos(bx) + (e^ax)(A sin(bx) + B cos(bx))'
=(a*e^ax)*(A sin(bx) + B cos(bx)) + (e^ax)(bA cos(bx) - bB sin(bx))
this simplifies to:
(e^ax)*((aA-bB) sin(bx) + (aB+bA) cos(bx))
We want this derivative to be equal to e^ax* sin(bx)

So this means that:
(aA - bB) = 1
(aB + bA) = 0
Now we can use algebra to solve for A and B. Try it!
(Hint: try solving the second equation for A, then plugging the result in for A in the first equation.)

Note that part b will be exactly the same, except this time we'll have:
(aA - bB) = 0
(aB + bA) = 1

2007-01-17 16:02:15 · answer #2 · answered by Anonymous · 0 0

By parts: ∫u dv = uv - ∫v du There is no immediated use of the formula, since you don't have a convenient substitution, however if we express all in terms of y: If y = cos^(-1) 2x cos y = 2x So: - sin y * dy/dx = 2 -1/2* siny dy = dx Substituting into your integral interms of y: ∫ y*-1/2* siny dy -1/2∫ sin y * ydy If u = y du = dy If dv = sin y dy v = - cos y In terms of y your integral is: - cos y * y - ∫(-cos y *dy = - y cos y + sin y + C = - 2x cos ^(-1) 2x + √(1 - 4x^2) + C

2016-05-24 01:16:43 · answer #3 · answered by ? 4 · 0 0

Yes its noit

2007-01-17 09:43:28 · answer #4 · answered by Anonymous · 0 0

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