Definite Integration Question - None

Question

Given $ I = \int_0^{\pi/2} \dfrac{\sin^3 x}{\sin x + \cos x}\,dx $. Find the value of $I$.

Accepted Answer

Step 1: Let I = ∫₀^(π/2) sin³x/(sin x + cos x) dx

Step 2: Using the property ∫₀^(π/2) f(x)dx = ∫₀^(π/2) f(π/2 - x)dx, substitute x → π/2 - x:

I = ∫₀^(π/2) sin³(π/2 - x)/(sin(π/2 - x) + cos(π/2 - x)) dx = ∫₀^(π/2) cos³x/(cos x + sin x) dx

Step 3: Add both expressions:

2I = ∫₀^(π/2) [sin³x + cos³x]/(sin x + cos x) dx

Step 4: Factor the numerator using a³ + b³ = (a + b)(a² - ab + b²):

sin³x + cos³x = (sin x + cos x)(sin²x - sin x cos x + cos²x) = (sin x + cos x)(1 - sin x cos x)

Step 5: Simplify:

2I = ∫₀^(π/2) (1 - sin x cos x) dx = ∫₀^(π/2) dx - ∫₀^(π/2) sin x cos x dx

Step 6: Evaluate:

2I = [x]₀^(π/2) - ½∫₀^(π/2) sin 2x dx = π/2 - ½[-cos 2x/2]₀^(π/2) = π/2 - ¼[(-1) - (1)] = π/2 + 1/2

Step 7: Therefore:

∴ I = (π + 1)/4 or (π/4 + 1/2) = B

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