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प्रश्न
` ∫ cos 3x cos 4x` dx
बेरीज
उत्तर
\[\int\text{cos 4x }\text{cos 3x dx}\]
` = 1/2 ∫ 2 cos 4x cos 3x dx `
\[ = \frac{1}{2}\int\left[ \text{cos} \left( 4x + 3x \right) + \text{cos }\left( 4x - 3x \right) \right]dx \left[ \therefore \text{2 }\text{cos A }\text{cos B} = \text{cos} \left( A + B \right) + \text{cos }\left( A - B \right) \right]\]
\[ = \frac{1}{2}\int\left( \text{cos} \left( 7x \right) + \text{cos x} \right) dx\]
\[ = \frac{1}{2}\left[ \frac{\sin 7x}{7} + \sin x \right] + C\]
\[ = \frac{1}{14}\sin 7x + \frac{1}{2}\sin x + C\]
\[ = \frac{1}{2}\int\left[ \text{cos} \left( 4x + 3x \right) + \text{cos }\left( 4x - 3x \right) \right]dx \left[ \therefore \text{2 }\text{cos A }\text{cos B} = \text{cos} \left( A + B \right) + \text{cos }\left( A - B \right) \right]\]
\[ = \frac{1}{2}\int\left( \text{cos} \left( 7x \right) + \text{cos x} \right) dx\]
\[ = \frac{1}{2}\left[ \frac{\sin 7x}{7} + \sin x \right] + C\]
\[ = \frac{1}{14}\sin 7x + \frac{1}{2}\sin x + C\]
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