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3.4 · Q167

Q.Evaluate: ∫12sin⁡x+sin⁡2x dx\int \frac{1}{2\sin x+\sin 2x}\,dx

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Since sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x, 2sin⁡x+sin⁡2x=2sin⁡x(1+cos⁡x)2\sin x+\sin2x=2\sin x(1+\cos x). Multiply numerator and denominator by sin⁡x\sin x: sin⁡x2sin⁡2x(1+cos⁡x)2\dfrac{\sin x}{2\sin^2x(1+\cos x)^2} (using sin⁡2x=(1−cos⁡x)(1+cos⁡x)\sin^2x=(1-\cos x)(1+\cos x), one factor of 1+cos⁡x1+\cos x cancels into the earlier 1+cos⁡x1+\cos x giving (1+cos⁡x)2(1+\cos x)^2 overall) =sin⁡x2(1−cos⁡x)(1+cos⁡x)2=\dfrac{\sin x}{2(1-\cos x)(1+\cos x)^2}. Let t=cos⁡xt=\cos x, dt=−sin⁡x dxdt=-\sin x\,dx: integral becomes −12∫dt(1−t)(1+t)2-\frac12\int\dfrac{dt}{(1-t)(1+t)^2}. Write 1(1−t)(1+t)2=A1−t+B1+t+C(1+t)2\dfrac{1}{(1-t)(1+t)^2}=\dfrac{A}{1-t}+\dfrac{B}{1+t}+\dfrac{C}{(1+t)^2}, so 1=A(1+t)2+B(1−t)(1+t)+C(1−t)1=A(1+t)^2+B(1-t)(1+t)+C(1-t). Put t=1t=1: A=14A=\frac14. Put t=−1t=-1: C=12C=\frac12. Comparing t2t^2 coefficients: 0=A−B⇒B=140=A-B \Rightarrow B=\frac14 (check at t=0t=0: $1 …

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