Question

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

Answer 1

\[ \text{ Let  I} = \int\sqrt{\sin x} \cdot \cos^3 \text{ x  dx }\]
\[ = \int\sqrt{\sin x} \cdot \left( \cos^2 x \right) \cdot \text{ cos  x  dx }\]
\[ = \int\sqrt{\sin x} \left( 1 - \sin^2 x \right) \cdot \text{ cos  x  dx}\]
\[\text{ Putting  sin x} = t\]
\[ \Rightarrow \text{ cos x  dx }= dt\]
\[ \therefore I = \int\sqrt{t} \left( 1 - t^2 \right) \cdot dt\]
\[ = \int t^\frac{1}{2} dt - \int t^\frac{1}{2} \cdot t^2 dt\]
\[ = \int t^\frac{1}{2} dt - \int t^\frac{5}{2} dt\]
\[ = \frac{t^\frac{3}{2}}{\frac{3}{2}} - \frac{t^\frac{7}{2}}{\frac{7}{2}} + C\]
\[ = \frac{2}{3} t^\frac{3}{2} - \frac{2}{7} t^\frac{7}{2} + C\]
\[ = \frac{2}{3} \text{ sin }^\frac{3}{2} \text{ x }- \frac{2}{7} \text{ sin }^\frac{7}{2} \text{ x }+ C ..........\left[ \because t = \text{ sin x }\right]\]