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
= $\frac{1}{2}$ ∫ $\frac{\mathit{s}\mathit{i}\mathit{n}2\mathit{x}}{2}$ dx - $\frac{1}{2}$ ∫ sin x cos 5x
= $\frac{1}{4}|\frac{-\mathit{c}\mathit{o}\mathit{s}2\mathit{x}}{2}|$ - $\frac{1}{2}\int \frac{1}{2}$ sin x + 5x + sin x - 5x dx
= $\frac{\mathit{c}\mathit{o}\mathit{s}2\mathit{x}}{8}$ - $\frac{1}{4}$ ∫ (sin 6x + sin (-4x) dx
= $\frac{-\mathit{c}\mathit{o}\mathit{s}2\mathit{x}}{8}$ - $\frac{1}{4}|\frac{-\mathit{c}\mathit{o}\mathit{s}6\mathit{x}}{6}+\frac{\mathit{c}\mathit{o}\mathit{s}4\mathit{x}}{4}|$ + C
= $\frac{-\mathit{c}\mathit{o}\mathit{s}2\mathit{x}}{8}$ - $\frac{1}{8}|\frac{-\mathit{c}\mathit{o}\mathit{s}6\mathit{x}}{3}+\frac{\mathit{c}\mathit{o}\mathit{s}4\mathit{x}}{2}|$ + C
= $\frac{-6\mathit{c}\mathit{o}\mathit{s}2\mathit{x}}{48}$ - $\frac{1}{8}|\frac{-2\mathit{c}\mathit{o}\mathit{s}6\mathit{x}+3\mathit{c}\mathit{o}\mathit{s}4\mathit{x}}{6}|$ + C
= $\frac{1}{48}$ [2 cos 6x - 3 cos 4x - 6 cos 2x] + C
OR
Let $\frac{2}{\left(1-\mathit{x}\right)\left(1+{\mathit{x}}^2\right)}$ = $\frac{\mathit{A}}{1-\mathit{x}}$ + $\frac{\mathit{B}\mathit{x}+\mathit{C}}{1+{\mathit{x}}^2}$
2 = A $\left(1+{\mathit{x}}^2\right)$ + Bx + X (1 - x)
2 = A + $\mathit{A}{\mathit{x}}^2$ + Bx - $\mathit{B}{\mathit{x}}^2$ + C - Cx
Equating the coefficient of ${\mathit{x}}^2$, 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
∴ $\frac{2}{\left(1-\mathit{x}\right)\left(1+{\mathit{x}}^2\right)}$ = $\frac{1}{1-\mathit{x}}$ + $\frac{\mathit{x}+1}{1+{\mathit{x}}^2}$
⇒ ∫ $\frac{2}{\left(1-\mathit{x}\right)\left(1+{\mathit{x}}^2\right)}$ dx = ∫ $\frac{1}{1-\mathit{x}}$ dx + ∫ $\frac{\mathit{x}}{1+{\mathit{x}}^2}$ dx + ∫ $\frac{1}{1+{\mathit{x}}^2}$ dx
= - ∫ $\frac{1}{\mathit{x}-1}$ dx + $\frac{1}{2}$ ∫ $\frac{2\mathit{x}}{1+{\mathit{x}}^2}$ dx + ∫ $\frac{1}{1+{\mathit{x}}^2}$ dx
= - log |x - 1| + $\frac{1}{2}$ log $\left|1+{\mathit{x}}^2\right|$ + $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ x + C