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3.4 · Q166

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

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Since sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x, sin⁡x+sin⁡2x=sin⁡x(1+2cos⁡x)\sin x+\sin2x=\sin x(1+2\cos x). Multiply numerator and denominator by sin⁡x\sin x: 1sin⁡x(1+2cos⁡x)=sin⁡xsin⁡2x(1+2cos⁡x)=sin⁡x(1−cos⁡2x)(1+2cos⁡x)\dfrac{1}{\sin x(1+2\cos x)}=\dfrac{\sin x}{\sin^2x(1+2\cos x)}=\dfrac{\sin x}{(1-\cos^2x)(1+2\cos x)}. Let t=cos⁡xt=\cos x, dt=−sin⁡x dxdt=-\sin x\,dx: the integral becomes −∫dt(1−t)(1+t)(1+2t)-\int\dfrac{dt}{(1-t)(1+t)(1+2t)}. Write 1(1−t)(1+t)(1+2t)=A1−t+B1+t+C1+2t\dfrac{1}{(1-t)(1+t)(1+2t)}=\dfrac{A}{1-t}+\dfrac{B}{1+t}+\dfrac{C}{1+2t}, so 1=A(1+t)(1+2t)+B(1−t)(1+2t)+C(1−t)(1+t)1=A(1+t)(1+2t)+B(1-t)(1+2t)+C(1-t)(1+t). Put t=1t=1: A=16A=\frac16. Put t=−1t=-1: B=−12B=-\frac12. Put t=−12t=-\frac12: C=43C=\frac43 (check at t=0t=0: 1=A+B+C=16−12+43=11=A+B+C=\frac16-\frac12+\frac43=1). So $-\int\left(\frac{ …

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