Question

\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]

Answer 1

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