Question

\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]

Answer 1

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