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प्रश्न
` ∫ sin 4x cos 7x dx `
योग
उत्तर
` ∫ sin 4x cos 7x dx `
\[ = \frac{1}{2}\int2 \cos 7x \sin \text{4x dx}\]
\[ = \frac{1}{2}\int\left[ \text{sin }\left( 7x + 4x \right) - \text{sin} \left( 7x - 4x \right) \right]dx \left[ \therefore 2 \cos A \sin B = \sin \left( A + B \right) - \sin \left( A - B \right) \right]\]
\[ = \frac{1}{2}\int\left( \text{sin }\left( \text{11x} \right) - \text{sin} \left( 3x \right) \right) dx\]
\[ = \frac{1}{2}\left[ - \frac{\text{cos} \left( 11x \right)}{11} + \frac{\text{cos} \left( 3x \right)}{3} \right] + C\]
\[ = - \frac{\text{cos} \left( 11x \right)}{22} + \frac{\text{cos} \left( 3x \right)}{6}\]
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