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प्रश्न
\[\int x \cos^3 x\ dx\]
बेरीज
उत्तर
Let I=\[\int x \cos^3 x\ dx\]
\[\text{ As we know }, \]
\[\cos 3x = 4 \cos^3 x - 3\cos x\]
\[ \Rightarrow \cos^3 x = \frac{1}{4}\left( \cos 3x + 3\cos x \right)\]
\[\cos 3x = 4 \cos^3 x - 3\cos x\]
\[ \Rightarrow \cos^3 x = \frac{1}{4}\left( \cos 3x + 3\cos x \right)\]
\[\therefore I = \frac{1}{4}\int x . \left( \cos 3x + 3 \cos x \right)dx\]
\[ = \frac{1}{4}\int x_I . \text{ cos}_{II} \left( \text{ 3x }\right) dx + \frac{3}{4} \int x_I . \cos x_{II} \text{ dx }\]
\[ = \frac{1}{4}\left[ x . \int\text{ cos 3x dx }- \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos 3x dx }\right\}dx \right] + \frac{3}{4}\left[ x\int\cos x - \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos x dx }\right\}dx \right]\]
\[ = \frac{1}{4}\left[ x . \frac{\sin 3x}{3} - \int1 . \frac{\sin 3x}{3}dx \right] + \frac{3}{4}\left[ x\left( \sin x \right) - \int1 . \text{ sin x dx } \right]\]
\[ = \frac{x \sin 3x}{12} + \frac{\cos 3x}{36} + \frac{3}{4}x \sin x + \frac{3}{4}\cos x + C\]
\[ = \frac{1}{4}\int x_I . \text{ cos}_{II} \left( \text{ 3x }\right) dx + \frac{3}{4} \int x_I . \cos x_{II} \text{ dx }\]
\[ = \frac{1}{4}\left[ x . \int\text{ cos 3x dx }- \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos 3x dx }\right\}dx \right] + \frac{3}{4}\left[ x\int\cos x - \int\left\{ \frac{d}{dx}\left( x \right) . \int\text{ cos x dx }\right\}dx \right]\]
\[ = \frac{1}{4}\left[ x . \frac{\sin 3x}{3} - \int1 . \frac{\sin 3x}{3}dx \right] + \frac{3}{4}\left[ x\left( \sin x \right) - \int1 . \text{ sin x dx } \right]\]
\[ = \frac{x \sin 3x}{12} + \frac{\cos 3x}{36} + \frac{3}{4}x \sin x + \frac{3}{4}\cos x + C\]
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