A Advanced Maths Integral?

Question

G'day.

Could someone please try and do this for me... Thanks

Evaluate: The Integral of; (sin x / (cos^2 x - 5cos x+4))

If you can, try and evaluate using partial fractions. But any standard Integral method that a Senior Maths student should know will be fine.

Thanks again.

Answer 1

∫ [sinx / (cos²x - 5cosx + 4)] dx =

note that your integrand includes both the function cosx and the opposite of its derivative, that is sinx; thus let cosx = u

differentiate both sides:

d(cosx) = du →

(-sinx) dx = du  Ã¢Â†Â’

sinx dx = - du

thus, substituting, you get:

∫ (sinx dx) / (cos²x - 5cosx + 4) = ∫ (- du) / (u² - 5u + 4) =

∫ [- 1 / (u² - 5u + 4)] du =

factor the denominator completely:

∫ [- 1 / (u² - u - 4u + 4)] du =

∫ {- 1 / [u(u - 1) - 4(u - 1)]} du =

∫ {- 1 / [(u - 1)(u - 4)]} du =

now set partial fraction decomposition:

- 1 / [(u - 1)(u - 4)] = A/(u - 1) + B/(u - 4) →

- 1 / [(u - 1)(u - 4)] = [A(u - 4) + B(u - 1)] / [(u - 1)(u - 4)] →

- 1 / [(u - 1)(u - 4)] = (Au - 4A + Bu - B) / [(u - 1)(u - 4)] →

- 1 / [(u - 1)(u - 4)] = [(A + B)u + (- 4A - B)] / [(u - 1)(u - 4)] →

thus, equating the numerators, you get:

| A + B = 0 → A = - B → A = 1/3

| - 4A - B = - 1 → - 4(- B) - B = - 1 → 4B - B = - 1 → 3B = - 1 → B = - 1/3

therefore (see above):

- 1 / [(u - 1)(u - 4)] = A/(u - 1) + B/(u - 4) →

- 1 / [(u - 1)(u - 4)] = (1/3)/(u - 1) - (1/3)/(u - 4)

your integral becoming:

∫ {- 1 / [(u - 1)(u - 4)]} du = ∫ {[(1/3)/(u - 1)] - [(1/3)/(u - 4)]} du =

breaking it up and taking the constants out,

(1/3) ∫ [1 /(u - 1)] du - (1/3) ∫ [1 /(u - 4)] du =

(1/3) ln | u - 1 | - (1/3) ln | u - 4 | + C =

(1/3) (ln | u - 1 | - ln | u - 4 |) + C =

according to log properties:

(1/3) ln | (u - 1) /(u - 4) | + C

finally, substituting back u = cosx, you get:

∫ [sinx / (cos²x - 5cosx + 4)] dx = (1/3) ln | (cosx - 1) /(cosx - 4) | + C

I hope it helps...

Bye!