It is known that, si A sin B = $\frac{1}{2}$ cos A - B - cos A + B ∴ ∫ sin x sin 2x sin 3x dx = ∫ |sin x × $\frac{1}{2}$ cos 2x - 3x - cos 2x + 3x| = $\frac{1}{2}$ ∫ sin x cos (-x) - sin x cos 5x dx = $\frac{1}{2}$ ∫ sin x cos x - sin x cos 5x dx = $\frac{1}{2}$ ∫ $\frac{\sin 2x}{2}$ dx - $\frac{1}{2}$ ∫ sin x cos 5x = $\frac{1}{4} \left| \frac{-\cos 2x}{2} \right|$ - $\frac{1}{2} \int \frac{1}{2}$ sin x + 5x + sin x - 5x dx = $\frac{\cos 2x}{8}$ - $\frac{1}{4}$ ∫ (sin 6x + sin (-4x) dx = $\frac{-\cos 2x}{8}$ - $\frac{1}{4} \left| \frac{-\cos 6x}{6} + \frac{\cos 4x}{4} \right|$ + C = $\frac{-\cos 2x}{8}$ - $\frac{1}{8} \left| \frac{-\cos 6x}{3} + \frac{\cos 4x}{2} \right|$ + C = $\frac{-6\cos 2x}{48}$ - $\frac{1}{8} \left| \frac{-2\cos 6x + 3\cos 4x}{6} \right|$ + C = $\frac{1}{48}$ [2 cos 6x - 3 cos 4x - 6 cos 2x] + C OR Let $\frac{2}{(1-x)(1+x^2)}$ = $\frac{A}{1-x}$ + $\frac{Bx + C}{1+x^2}$ 2 = A $(1+x^2)$ + Bx + X (1 - x) 2 = A + $Ax^2$ + Bx - $Bx^2$ + C - Cx Equating the coefficient of $x^2$, x, and constant term, we obtain A − B = 0 B − C = 0 A + C = 2 On solving these equations, we obtain A = 1, B = 1, and C = 1 ∴ $\frac{2}{(1-x)(1+x^2)}$ = $\frac{1}{1-x}$ + $\frac{x+1}{1+x^2}$ ⇒ ∫ $\frac{2}{(1-x)(1+x^2)}$ dx = ∫ $\frac{1}{1-x}$ dx + ∫ $\frac{x}{1+x^2}$ dx + ∫ $\frac{1}{1+x^2}$ dx = - ∫ $\frac{1}{x-1}$ dx + $\frac{1}{2}$ ∫ $\frac{2x}{1+x^2}$ dx + ∫ $\frac{1}{1+x^2}$ dx = - log |x - 1| + $\frac{1}{2}$ log $\left| 1+x^2 \right|$ + $\tan^{-1}$ x + C