Question

Integration of sin-1 root x - cos-1 root x /sin-1 rootx + cos-1 root x

Answer 1

Step-by-step explanation:

Given Integration of sin-1 root x - cos-1 root x /sin-1 root x + cos-1 root x

• So we have ∫ sin^-1 √x – cos ^-1 √x / sin^-1 √x + cos ^-1 √x
• So we have sin^-1 theta + cos^-1 theta = π / 2
• So ∫ sin^-1√x – (π / 2 – sin^-1 √x) / π / 2 dx
•  So we get 2/π ∫ (2 sin^-1√x – π/2 ) dx
• I = 4/π ∫ sin^-1 √x / I1 dx – 2/π x π/2 ∫ dx
• So I1 = ∫sin^-1 √x dx  
• Now let √x = t
• Differentiating we get
•  1/2√x dx = dt
• So dx = 2t dt
•          = 2 ∫ t sin^-1 t dt
•          = 2 sin^-1 t ∫t dt – 2 ∫ (d sin^-1 t) / dt ∫t dt) dt
•          = 2 t^2 / 2 sin^-1 t – 2 ∫ 1 / √1 – t^2 x t^2 / 2 dt
•          = t^2 sin^-1 t + ∫ - t^2 / √1 – t^2 dt
•      Add and subtract 1 we get
•            t^2 sin^-1 t + ∫ 1 - t^2 + 1 / √1 – t^2 dt
•      so t^2 sin^-1 t + ∫1 – t^2 / √1 – t^2 dt - ∫ dt / √1 – t^2
•      So t^2 sin^-1 t – sin^-1 t + ∫√1 – t^2 dt
• So [∫√a^2 – x^2 dx = ½ x √a^2 – x^2 + a^2 / 2 sin^-1 (x/a) + C]
• So t^2 sin^-1 t – sin^-1 t + t √1 – t^2 / 2 + ½ sin^-1 t + C
• So t^2 sin^-1 t + t √1 – t^2 / 2 – ½ sin^-1 t + C1
• So I1 = (√x)^2 = x
• So I1 = x sin^-1 √x + √x √1 – x /2 – ½ sin^-1√x + C1
• So I = 4/π I1
•       = 4/π (x sin^-1 √x + √1 – x^2 / 2 – ½ sin^-1 √x + C1) - ∫dx
•       = 4/π (x sin^-1 √x + √1 – x^2 / 2 – ½ sin^-1√x + C1) – x + C2
•     = 4x /π sin^-1√x + 2/ π√1 – x^2 – 2/π sin^-1 √x – x + C   (4/π C1 + C2 = C)

Reference link will be

https://brainly.in/question/4272483

Answer 2

Answer:

upper one is correct answer