∫ $\frac{\sin(x-a)}{\sin(x+a)}$ dx Let (x + a) = t ⇒ dx = dt ∴ I = ∫ $\frac{\sin t - 2a}{\sin t}$ dt = ∫ $\frac{\sin t \cos 2a - \cos t \sin 2a}{\sin t}$ dt = ∫ cos 2a - cot t sin 2a dt = cos 2a t - sin 2a log |sin t| + C = cos 2a x + a - sin 2a log |sin (x + a)| + C OR ∫ $\frac{5x-2}{1+2x+3x^2}$ dx = 5 ∫ $\frac{x - \frac{2}{5}}{1+2x+3x^2}$ dx = $\frac{5}{6}$ ∫ $\frac{6x - \frac{12}{5}}{1+2x+3x^2}$ dx = $\frac{5}{6}$ ∫ $\frac{6x + 2 - \frac{12}{5} - 2}{1+2x+3x^2}$ dx = $\frac{5}{6}$ ∫ $\frac{6x + 2 - \frac{22}{5}}{1+2x+3x^2}$ dx = $\frac{5}{6}$ ∫ $\frac{6x+2}{1+2x+3x^2}$ - $\frac{5}{6} \times \frac{22}{5}$ ∫ $\frac{1}{3\left[\left(x+\frac{1}{3}\right)^2 + \frac{2}{9}\right]}$ dx = $\frac{5}{6}$ log |1 + 2x + $3x^2$| - $\frac{11}{9}$ ∫ $\frac{1}{\left(x+\frac{1}{3}\right)^2 + \frac{2}{9}}$ dx = $\frac{5}{6}$ log |1 + 2x + $3x^2$| - $\frac{11}{9} \times \frac{3}{\sqrt{2}}$ $\frac{\tan^{-1}\left(x+\frac{1}{3}\right)}{\frac{\sqrt{2}}{3}}$ + C = $\frac{5}{6}$ log |1 + 2x + $3x^2$| - $\frac{11}{3\sqrt{2}}$ × $\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)$ + C