Question

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

Answer 1

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