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प्रश्न
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
उत्तर
\[Let, I = \int_0^\frac{\pi}{4} \sin2x \sin3x d x ..................(1)\]
\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} 2\cos2x\frac{\cos3x}{3}dx\]
\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} + \frac{4}{9} \int_0^\frac{\pi}{4} \sin2x \sin3x d x\]
\[ \Rightarrow I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} + \frac{4}{9}I ..............\left[From (1) \right]\]
\[ \Rightarrow \frac{5}{9}I = \left[ - \sin2x\frac{\cos3x}{3} \right]_0^\frac{\pi}{4} + \frac{2}{3} \left[ \cos2x\frac{\sin3x}{3} \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow \frac{5}{9}I = \frac{1}{3\sqrt{2}} + 0\]
\[ \Rightarrow \frac{5}{9}I = \frac{1}{3\sqrt{2}}\]
\[ \therefore I = \frac{3}{5\sqrt{2}}\]
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