- 2067 Asar
Show ∫ sin²x/(sin x+cos x) dx
Short answer
Replace x by π/2−x and add to obtain 2I = ∫₀^{π/2} dx/(sin x+cos x). Write sin x+cos x = √2 sin(x+π/4), integrate csc to get 2I = (2/√2) ln(√2+1). Hence I = (1/√2) ln(√2+1).
Prove that ∫₀^{π/2} sin²x dx/(sin x + cos x) = (1/√2) log(√2+1).
Let I = ∫₀^(π/2) sin²x/(sin x + cos x) dx. Replacing x by π/2 − x:
I = ∫₀^(π/2) cos²x/(sin x + cos x) dx
Add the two forms; since sin²x + cos²x = 1:
2I = ∫₀^(π/2) dx/(sin x + cos x) = K
Evaluate K. sin x + cos x = √2 sin(x + π/4), so
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)]
= (1/√2) ln[(√2+1)/(√2−1)]
= (1/√2) · 2 ln(√2 + 1)
So 2I = (2/√2) ln(√2 + 1), and
I = (1/√2) ln(√2 + 1). Proved.
✓ Answer checked by Claude Opus 5.5 (high effort)
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