Question

\[\int\frac{\cos^5 x}{\sin x} \text{ dx }\]

Answer 1

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