intergral help ? plz show the working?

Question

integrate

(3+2COSx)/((2+3COSx)^2)

Answer 1

This is a difficult integral, but it has a simple solution:

sin(x) / (2+3cos(x))

The best way to go about this is to note that you should separate the fraction into two integrals, using partial fractions (with the substitution u=3+2cos(x). The denominator then becomes 2(u-1)(u-2), and don't forget that du = -sqrt((5-u)(u-1)) dx. This of course gives you another difficult integral, but one that can be tediously worked out using partial fractions and arctans.

Steve

Answer 2

We can first convert this into another form similar to partial fractions. Let u = cos(x) {but remember we're not ACTUALLY performing substitution with respect to integration}. Then, we can develop this partial fractions decomposition:

(3 + 2u) / (2 + 3u)^2 = A/(2 + 3u) + B/(2 + 3u)^2

3 + 2u = A(2 + 3u) + B
3 + 2u = 2A + 3Au + B
3 + 2u = (2A + B) + 3Au

Therefore,
3A = 2
2A + B = 2, B = 2 - 2A

A = 2/3, B = 2 - 2(2/3) = 2 - 4/3 = 2/3

Therefore, our partial fractions decomposition is

(2/3) (1 / [2 + 3u] ) + (2/3) (2 + 3u)^2

Substituting u = cos(x) back, we have

(2/3) (1 / [2 + 3cosx] ) + (2/3) (1 / [2 + 3cos(x)]^2

So we want to find the integral of that above. We can separate it into two integrals, though.

(2/3) * Integral (1 / [2 + 3cosx])dx +
(2/3) * Integral (1 / [2 + 3cosx]^2 dx

This integral is tough. Not sure where to go from here.

Answer 3

Are there bounds of integration for this integral, or are you asking for the anti-derivative? Because for certain bounds of integration, this should be an easy problem if you apply the residue theorem.

Answer 4

Regroup the numerator:

3+2cosx
= 3(cosx)^2 + 3(sinx)^2 + 2cosx
= cosx(2+3cosx) + 3(sinx)^2
= (sinx)'(2+3cosx) - sinx(2+3cosx)'

Therefore,
ʃ(3+2COSx)/((2+3COSx)^2)
= ʃd[sinx/(2+3cosx)]
= sinx/(2+3cosx) + c

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a_math_gu,

You just copied my solution, didn't you?

Answer 5

The integrand is [3+2cos(x)]/ [(2+3cos(x))^2] = [3cos(x)^2 + 3sin(x)^2 +2cos(x)]/ [(2+3cos(x))^2] = {(2+3cos(x))*cos(x) +3sin(x)^2}/ [(2+3cos(x))^2] = (2+3cos(x))*cos(x)/ [(2+3cos(x))^2] +3sin(x)^2/ [(2+3cos(x))^2] = cos(x)/ (2+3cos(x) + 3sin(x)^2/ [(2+3cos(x))^2] = cos(x)/ (2+3cos(x)) + sin(x)* -1/(2+3cos(x))^2 * -3sin(x) = {sin(x)}' * 1/(2+3cos(x)) + sin(x)* {1/(2+3cos(x))}' = {sin(x) * 1/(2+3cos(x))}' by the product rule.

So integral = sin(x)/(3+2cos(x)) + C

OK now?

My suggestion (again) is software for harder integrals.

Sigh.