Question

\[\int\frac{\sin^3 x}{\sqrt{\cos x}} dx\]

Answer 1

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