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प्रश्न
\[\int\frac{\cos^7 x}{\sin x} dx\]
बेरीज
उत्तर
\[\text{ Let I } = \int\frac{\cos^7 x}{\sin x}dx\]
\[ = \int\frac{\cos^6 x \cdot \cos x dx}{\sin x}\]
\[ = \int\frac{\left( \cos^2 x \right)^3 \cdot \cos x}{\sin x}dx\]
\[ = \int\frac{\left( 1 - \sin^2 x \right)^3 \cdot \cos x}{\sin x}dx\]
\[\text{ Let sin x} = t\]
\[ \Rightarrow \text{ cos x dx } = dt\]
\[ \therefore I = \int\frac{\left( 1 - t^2 \right)^3}{t}dt\]
\[ = \int\left( \frac{1 - t^6 - 3 t^2 + 3 t^4}{t} \right)\text{ dt }\]
\[ = \int\left( \frac{1}{t} - t^5 - 3t + 3 t^3 \right)\text{ dt}\]
\[ = \text{ ln }\left| t \right| - \frac{t^6}{6} - \frac{3 t^2}{2} + \frac{3 t^4}{4} + C\]
\[ = \text{ ln }\left| \sin x \right| - \frac{\sin^6 x}{6} - \frac{3 \sin^2 x}{2} + \frac{3}{4} \sin^4 x + C ...............\left( \because t = \sin x \right)\]
\[ = \int\frac{\cos^6 x \cdot \cos x dx}{\sin x}\]
\[ = \int\frac{\left( \cos^2 x \right)^3 \cdot \cos x}{\sin x}dx\]
\[ = \int\frac{\left( 1 - \sin^2 x \right)^3 \cdot \cos x}{\sin x}dx\]
\[\text{ Let sin x} = t\]
\[ \Rightarrow \text{ cos x dx } = dt\]
\[ \therefore I = \int\frac{\left( 1 - t^2 \right)^3}{t}dt\]
\[ = \int\left( \frac{1 - t^6 - 3 t^2 + 3 t^4}{t} \right)\text{ dt }\]
\[ = \int\left( \frac{1}{t} - t^5 - 3t + 3 t^3 \right)\text{ dt}\]
\[ = \text{ ln }\left| t \right| - \frac{t^6}{6} - \frac{3 t^2}{2} + \frac{3 t^4}{4} + C\]
\[ = \text{ ln }\left| \sin x \right| - \frac{\sin^6 x}{6} - \frac{3 \sin^2 x}{2} + \frac{3}{4} \sin^4 x + C ...............\left( \because t = \sin x \right)\]
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