Question

\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]

Answer 1

\[Let, I = \int_0^\frac{\pi}{4} \sin2x \sin3x d x ..................(1)\]

\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} 2\cos2x\frac{\cos3x}{3}dx\]

\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} + \frac{4}{9} \int_0^\frac{\pi}{4} \sin2x \sin3x d x\]

\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} + \frac{4}{9}I ..............\left[From (1) \right]\]

\[ \Rightarrow \frac{5}{9}I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} \]

\[ \Rightarrow \frac{5}{9}I = \frac{1}{3\sqrt{2}} + 0\]

\[ \Rightarrow \frac{5}{9}I = \frac{1}{3\sqrt{2}}\]

\[ \therefore I = \frac{3}{5\sqrt{2}}\]