Question

Integrate: sinx/sin4x

Answer 1

let's write the denominator as sin[2(2x)] and apply the double-angle formula sin(2θ) = 2sinθ cosθ: sin[2(2x)] = 2sin(2x) cos(2x) = let's apply the double-angle formulae sin(2θ) = 2sinθ cosθ and cos(2θ) = cos²θ - sin²θ: 2(2sinx cosx) (cos²x - sin²x) = 4sinx cosx (cos²x - sin²x) then the integral becomes: ∫ {sinx /[4sinx cosx (cos²x - sin²x)]} dx = simplifying to: ∫ {1 /[4cosx (cos²x - sin²x)]} dx = let's divide an multiply by cosx: ∫ {1 /[4cos²x (cos²x - sin²x)]} cosx dx = let's rewrite the denominator in terms of sin²x, replacing each cos²x with 1 - sin²x: ∫ {1 /[4(1 - sin²x) (1 - sin²x - sin²x)]} cosx dx = (factoring 1/4 out) (1/4) ∫ {1 /[(1 - sin²x) (1 - 2sin²x)]} cosx dx = let: sinx = u (differentiating both sides) d(sinx) = du cosx dx = du yielding by substitution: (1/4) ∫ {1 /[(1 - sin²x) (1 - 2sin²x)]} cosx dx = (1/4) ∫ {1 /[(1 - u²)(1 - 2u²)]} du let's factor the denominator and decompose the integrand into partial fractions: 1 /[(1 - u²)(1 - 2u²)] = 1 /{(1 - u²){1 - [(√2)u]²} } = 1 /{(1 + u)(1 - u) [1 + (√2)u][1 - (√2)u]} = A/(1 + u) + B/(1 - u) + C/[1 + (√2)u] + D/[1 - (√2)u] ((1 + u)(1 - u) [1 + (√2)u][1 - (√2)u]: common denominator) 1 /{(1 + u)(1 - u)[1 + (√2)u][1 - (√2)u]} = {A(1 - u)[1 + (√2)u][1 - (√2)u] + B(1 + u)[1 + (√2)u][1 - (√2)u] + C(1 + u)(1 - u)[1 - (√2)u] + D(1 + u)(1 - u)[1 + (√2)u]} /{(1 + u)(1 - u)[1 + (√2)u][1 - (√2)u]} (equating numerators) 1 = A(1 - u)(1 - 2u²) + B(1 + u)(1 - 2u²) + C(1 - u²)[1 - (√2)u] + D(1 - u²)[1 + (√2)u] 1 = A(1 - 2u² - u + 2u³) + B(1 - 2u² + u - 2u³) + C[1 - (√2)u - u² + (√2)u³] + D[1 + (√2)u - u² - (√2)u³] 1 = A - 2Au² - Au + 2Au³ + B - 2Bu² + Bu - 2Bu³ + C - (√2)Cu - Cu² + (√2)Cu³ + D + (√2)Du - Du² - (√2)Du³ 1 = [2A - 2B + (√2)C - (√2)D]u³ + (- 2A - 2B - C - D)u² + [- A + B - (√2)C + (√2)D]u + (A + B + C + D) (equating coefficients) 2A - 2B + (√2)C - (√2)D = 0 - 2A - 2B - C - D = 0 - A + B - (√2)C + (√2)D = 0 A + B + C + D = 1 (√2)A - (√2)B + C - (- 2A - 2B - C) = 0 2A + 2B + C + D = 0 → D = - 2A - 2B - C - A + B - (√2)C + (√2)(- 2A - 2B - C) = 0 A + B + C + (- 2A - 2B - C) = 1 (√2)A - (√2)B + C + 2A + 2B + C = 0 D = - 2A - 2B - C - A + B - (√2)C - 2(√2)A - 2(√2)B - (√2)C = 0 A + B + C - 2A - 2B - C = 1 → - A - B = 1 → A + B = - 1 (√2)A - (√2)B + 2A + 2B + 2C = 0 → A - B + (√2)A + (√2)B + (√2)C = 0 D = - 2A - 2B - C - A + B - 2(√2)A - 2(√2)B - 2(√2)C = 0 A = - B - 1 (- B - 1) - B + (√2)(- B - 1) + (√2)B + (√2)C = 0 D = - 2(- B - 1) - 2B - C - (- B - 1) + B - 2(√2)(- B - 1) - 2(√2)B - 2(√2)C = 0 A = - B - 1 - B - 1 - B - (√2)B - √2 + (√2)B + (√2)C = 0 D = 2B + 2 - 2B - C = - C + 2 B + 1 + B + 2(√2)B + 2√2 - 2(√2)B - 2(√2)C = 0 A = - B - 1 - 2B - 1 - √2 + (√2)C = 0 D = - C + 2 2B + 1 + 2√2 - 2(√2)C = 0 → 2B = - 1 - 2√2 + 2(√2)C A = - B - 1 - [- 1 - 2√2 + 2(√2)C] - 1 - √2 + (√2)C = 0 D = - C + 2 2B = - 1 - 2√2 + 2(√2)C A = - B - 1 1 + 2√2 - 2(√2)C - 1 - √2 + (√2)C = 0 D = - C + 2 B = [- 1 - 2√2 + 2(√2)C] /2 A = - B - 1 √2 - (√2)C = 0 → 1 - C = 0 → - C = - 1 → C = 1 D = - C + 2 = - 1 + 2 = 1 B = [- 1 - 2√2 + 2(√2)C] /2 = [- 1 - 2√2 + 2(√2)1] /2 = (- 1 - 2√2 + 2√2) /2 = - 1/2 A = - B - 1 = - (-1/2) - 1 = (1/2) - 1 = - 1/2 obtaining: 1 /{(1 + u)(1 - u) [1 + (√2)u][1 - (√2)u]} = A/(1 + u) + B/(1 - u) + C/[1 + (√2)u] + D/[1 - (√2)u] = (-1/2)/(1 + u) + (-1/2)/(1 - u) + 1/[1 + (√2)u] + 1/[1 - (√2)u] the integral becomes: (1/4) ∫ {1 /[(1 - u²)(1 - 2u²)]} du = (1/4) ∫ {- [(1/2) /(1 + u)] - [(1/2) /(1 - u)] + {1 /[1 + (√2)u]} + {1 /[1 - (√2)u]} } du = (splitting into four integrals and factoring constants out) (1/4)(-1/2) ∫ [1 /(1 + u)] du + (1/4)(1/2) ∫ [- 1 /(1 - u)] du + (1/4) ∫ {1 /[1 + (√2)u]} du + (1/4) ∫ {1 /[1 - (√2)u]} du = - (1/8) ∫ d(1 + u) /(1 + u) + (1/8) ∫ d(1 - u) /(1 - u) + (1/4) ∫ {1 /[1 + (√2)u]} du + (1/4) ∫ {1 /[1 - (√2)u]} du = - (1/8) ln |1 + u| + (1/8) ln |1 - u| + (1/4) ∫ {1 /[1 + (√2)u]} du + (1/4) ∫ {1 /[1 - (√2)u]} du = let's divide and multiply the remaining integrals by √2 and - √2 respectively to make the numerator the derivative of the denominator: - (1/8) ln |1 + u| + (1/8) ln |1 - u| + (1/4)(1/√2) ∫ {(√2) /[1 + (√2)u]} du + (1/4)(- 1/√2) ∫ {(- √2) /[1 - (√2)u]} du = - (1/8) ln |1 + u| + (1/8) ln |1 - u| + [1/(4√2)] ∫ d[1 + (√2)u] /[1 + (√2)u] - [1/(4√2)] ∫ d[1 - (√2)u] /[1 - (√2)u] = - (1/8) ln |1 + u| + (1/8) ln |1 - u| + [1/(4√2)] ln |1 + (√2)u| - [1/(4√2)] ln |1 - (√2)u| + C = (1/8) (ln |1 - u| - ln |1 + u|) + [1/(4√2)] [ln |1 + (√2)u| - ln |1 - (√2)u|] + C = (applying logarithm properties) (1/8) ln |(1 - u) /(1 - u)| + [1/(4√2)] ln |[1 + (√2)u] /[1 - (√2)u]| + C finally let's substitute back sinx for u, ending with: (1/8) ln [(1 - sinx)/(1 + sinx)] + [1/(4√2)] ln |[1 + (√2)sinx] /[1 - (√2)sinx]| + C (given that - 1 < sinx < 1, the argument of the first logarithm is always positive, then the first absolute value is not needed any longer)