Question

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

Answer 1

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