Question

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

Answer 1

\[\int\frac{\cos^5 x}{\sin x}dx\]
\[ = \int \frac{\cos^4 x . \cos x}{\sin x}dx\]
\[ = \int\frac{\left( \cos^2 x \right)^2 . \cos x}{\sin x}dx\]
\[ = \int\frac{\left( 1 - \sin^2 x \right)^2 \times \cos x}{\sin x}dx\]
\[ = \int \frac{\left( 1 - \sin^4 x - 2 \sin^2 x \right)}{\sin x}\text{cos x dx}\]
\[\text{Let sin x }= t\]
\[ \Rightarrow \text{cos x dx} = dt\]
\[Now, \int \frac{\left( 1 - \sin^4 x - 2 \sin^2 x \right)}{\sin x}\text{cos x dx}\]
\[ = \int \frac{\left( 1 + t^4 - 2 t^2 \right)}{t}dt\]
\[ = \int\left( \frac{1}{t} + t^3 - 2t \right)dt\]
\[ = \text{log }\left|\text{ t} \right| + \frac{t^4}{4} - \frac{2 t^2}{2} + C\]
\[ = \text{log} \left|\text{ t }\right| + \frac{t^4}{4} - t^2 + C\]
\[ = \text{log }\left| \sin x \right| + \frac{\sin^4 x}{4} - \sin^2 x + C\]