Integral of (sin x+cos x)/(9+16 sin 2x)

Short answer

Substitute t = sin x – cos x, dt = (cos x + sin x)dx, sin 2x = 1 – t². Limits become -1 to 0. Integral = ∫_{-1}^0 dt/(25 – 16t²) = (1/20)ln 3.

Evaluate ∫₀^{π/4} (sin x + cos x)/(9 + 16 sin 2x) dx.

Let I = ∫₀^(π/4) (sin x + cos x)/(9 + 16 sin 2x) dx.

Put t = sin x − cos x. Then dt = (cos x + sin x) dx, and

t² = 1 − 2 sin x cos x = 1 − sin 2x, so sin 2x = 1 − t².

Limits: x = 0 → t = −1; x = π/4 → t = 0.

I = ∫_(−1)^0 dt / (9 + 16(1 − t²))
  = ∫_(−1)^0 dt / (25 − 16t²)
  = ∫_(−1)^0 dt / (5² − (4t)²)

Use ∫ dt/(c² − k²t²) = (1/(2ck)) ln((c + kt)/(c − kt)) with c = 5, k = 4:

I = (1/40) [ln((5 + 4t)/(5 − 4t))]_(−1)^0
  = (1/40) [ln 1 − ln(1/9)]
  = (1/40) ln 9
  = (1/20) ln 3

Answer: I = (1/20) ln 3 ≈ 0.0549.

✓ Answer checked by Claude Opus 5.5 (high effort)

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