Question

How can I COMPLETELY solve this system: sec(a) + cosec(b) = 2$\sqrt{2}$ sin(b) + cos(a) = $\sqrt{2}$

Answer 1

Answer:∫ csc[(3x)/2] sec(x/2) dx =

let's write it as:

∫ {1 /sin[(3x)/2]} [1 /cos(x/2)] dx =

∫ {1 /{sin[(3x)/2] cos(x/2)} } dx =

let's apply the product-to-sum formula sin α cosβ = [sin(α + β) + sin(α - β)] /2:

∫ {1 /{{sin{[(3x)/2] + (x/2)} + sin{[(3x)/2] - (x/2)}} /2} } dx =

∫ {1 /{{sin[(4x)/2] + sin[(2x)/2]} /2} } dx =

∫ {2 /[sin(2x) + sinx]} dx =

(applying the double-angle identity sin(2x) = 2sinx cosx)

∫ [2 /(2sinx cosx + sinx)] dx =

(factoring sinx out of the denominator)

∫ {2 /[sinx (2cosx + 1)]} dx =

let's divide and multiply by sinx:

∫ {2 /[sin²x (2cosx + 1)]} sinx dx =

let's replace sin²x with 1 - cos²x:

∫ {2 /[(1 - cos²x)(2cosx + 1)]} sinx dx =

let's factor the denominator completely:

∫ {2 /[(1² - cos²x)(2cosx + 1)]} sinx dx =

∫ {2 /[(1 + cosx)(1 - cosx)(2cosx + 1)]} sinx dx =

let:

cosx = u

let's differentiate both sides:

d(cosx) = du

- sinx dx = du

sinx dx = - du

yielding, by substitution:

∫ {2 /[(1 + cosx)(1 - cosx)(2cosx + 1)]} sinx dx = ∫ {2 /[(1 + u)(1 - u)(2u +  

1)]} (- du) =

(adjusting signs)

∫ {2 /[- (1 + u)(1 - u)(2u + 1)]} du =

∫ {2 /{(1 + u)[- (1 - u)] (2u + 1)} } du =

∫ {2 /[(u + 1)(u - 1)(2u + 1)]} du =

let's decompose into partial fractions:

2 /[(u + 1)(u - 1)(2u + 1)] = A/(u + 1) + B/(u - 1) + C/(2u + 1)

(letting (u + 1)(u - 1)(2u + 1) be the common denominator)

2 /[(u + 1)(u - 1)(2u + 1)] = [A(u - 1)(2u + 1) + B(u + 1)(2u + 1) + C(u + 1)(u -  

1)] /[(u + 1)(u - 1)(2u + 1)]

(equating numerators)

2 = A(u - 1)(2u + 1) + B(u + 1)(2u + 1) + C(u + 1)(u - 1)

2 = A(2u² + u - 2u - 1) + B(2u² + u + 2u + 1) + C(u² - 1²)

2 = A(2u² - u - 1) + B(2u² + 3u + 1) + C(u² - 1)

2 = 2Au² - Au - A + 2Bu² + 3Bu + B + Cu² - C

2 = (2A + 2B + C)u² + (- A + 3B)u + (- A + B - C)

(equating numerators)

2A + 2B + C = 0

- A + 3B = 0

- A + B - C = 2

2(3B) + 2B + C = 0

- A = - 3B → A = 3B

- (3B) + B - C = 2

6B + 2B + C = 0

A = 3B

- 2B - C = 2

8B + C = 0

A = 3B

- 2B - C = 2

C = - 8B

A = 3B

- 2B - (- 8B) = 2

C = - 8B

A = 3B

- 2B + 8B = 2

C = - 8B

A = 3B

6B = 2

C = - 8B = - 8(1/3) = - 8/3

A = 3B = 3(1/3) = 1

B = 2/6 = 1/3

yielding:

2 /[(u + 1)(u - 1)(2u + 1)] = A/(u + 1) + B/(u - 1) + C/(2u + 1) = 1/(u + 1) +  

(1/3)/(u - 1) + (- 8/3)/(2u + 1)

then the integral becomes:

∫ {2 /[(1 + u)(u - 1)(2u + 1)]} du = ∫ {[1 /(u + 1)] + [(1/3) /(u - 1)] + [(- 8/3) /(2u + 1)]} du =

(splitting into three integrals and factoring constants out)

∫ [1 /(u + 1)] du + (1/3) ∫ [1 /(u - 1)] du - (8/3) ∫ [1 /(2u + 1)] du =

(dividing and multiplying the last integral by 2 to make the numerator the derivative of the denominator)

∫ [1 /(u + 1)] du + (1/3) ∫ [1 /(u - 1)] du - (8/3)(1/2) ∫ 2 du /(2u + 1) =

ln |u + 1| + (1/3) ln |u - 1| - (4/3) ∫ d(2u + 1) /(2u + 1) =

(by the integration rule ∫ d[f(x)] /f(x) = ln |f(x)| + C)

ln |u + 1| + (1/3) ln |u - 1| - (4/3) ln |2u + 1| + C

let's substitute back cosx for u:

ln |cosx + 1| + (1/3) ln |cosx - 1| - (4/3) ln |2cosx + 1| + C =

since - 1 < cosx < 1

explanation: in (cosx + 1) + (1/3) ln (1 - cosx) - (4/3) ln |2cosx + 1| + C

Answer 2

${\red{\underline{\underline{\bold{Answer:-}}}}}$

see the attachment..

hope it helps you..