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Show ∫ x/(sin x+cos x) dx

Short answer

Use symmetry to obtain 2I = (π/2)∫_0^{π/2} dx/(sin x+cos x). Write denominator as √2 sin(x+π/4) and integrate to get √2 ln(√2+1), thus I = (π/(2√2)) ln(√2+1).

Show that ∫₀^{π/2} x/(sin x + cos x) dx = (π/(2√2)) log(√2+1).

Let I = ∫₀^(π/2) x/(sin x + cos x) dx. Using ∫₀^a f(x)dx = ∫₀^a f(a − x)dx with a = π/2 (sin and cos swap, the sum is unchanged):

I = ∫₀^(π/2) (π/2 − x)/(sin x + cos x) dx

Add: 2I = (π/2) ∫₀^(π/2) dx/(sin x + cos x).

Evaluate K = ∫₀^(π/2) dx/(sin x + cos x). Write sin x + cos x = √2 sin(x + π/4):

K = (1/√2) ∫₀^(π/2) cosec(x + π/4) dx
  = (1/√2) [ln tan((x + π/4)/2)]₀^(π/2)
  = (1/√2) [ln tan(3π/8) − ln tan(π/8)]

tan(3π/8) = √2 + 1 and tan(π/8) = √2 − 1, so the ratio is (√2 + 1)/(√2 − 1) = (√2 + 1)².

K = (1/√2) · 2 ln(√2 + 1) = √2 ln(√2 + 1).

Then 2I = (π/2) · √2 ln(√2 + 1), so I = (π√2/4) ln(√2 + 1).

I = (π/(2√2)) ln(√2 + 1). Shown.

✓ Answer checked by Claude Opus 5.5 (high effort)

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