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प्रश्न
उत्तर
\[\int \cos^3 \left( 3x \right) \text{ dx} \]
` \text{ As we know that cos 3A = 4 cos^{3} A - 3 cos A } `
\[ \Rightarrow \frac{\cos 3A + 3 \cos A}{4} = \cos^3 A\]
\[ \Rightarrow \int\left[ \frac{\cos \text{ 9x }+ 3 \cos 3x}{4} \right]dx\]
\[ \Rightarrow \frac{1}{4}\int\text{ cos 9x dx} + \frac{3}{4}\int\text{ cos 3x dx}\]
\[ \Rightarrow \frac{1}{4} \left[ \frac{\sin \text{ 9x}}{9} \right] + \frac{3}{4}\left[ \frac{\sin 3x}{3} \right] + C\]
\[ \Rightarrow \frac{\sin 9x}{36} + \frac{\sin 3x}{4} + C\]
\[ \Rightarrow \frac{1}{36}\left[ 3 \sin 3x - 4 \sin^{ 3} 3x \right] + \frac{\sin 3x}{4} + C................................. \left( \because \sin 3x = 3 \sin x - 4 \sin^3 x \right)\]
\[ \Rightarrow \frac{\sin 3x}{12} + \frac{\sin 3x}{4} - \frac{1}{9} \sin^3 3x + C\]
\[ \Rightarrow \frac{\sin 3x}{3} - \frac{\sin^3 3x}{9} + C\]
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