Question

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

Answer 1

\[\text{ Let, I } = \int \cos^3 \sqrt{x} \text{ dx } . . . . . \left( 1 \right)\]
\[\text{ Consider, }\sqrt{x} = t . . . . . \left( 2 \right)\]
\[\text{Differentiating both sides we get}, \]
\[\frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow dx = 2\sqrt{x} dt\]
\[ \Rightarrow dx = 2t dt\]
\[\text{ Therefore,} \left( 1 \right) \text{ becomes,} \]
\[I = \int \cos^3 \text{ t  2t  dt }\]
\[ = 2\int t  \text{ cos}^3\text{  t   dt}\]
\[ = 2\int \text{ t }\left( \frac{3\cos t + \cos3t}{4} \right) dt \left( \text{ Since,} \cos 3A = 4 \cos^3 A - 3\cos  A \right)\]
\[ = \frac{3}{2}\int \text{ t  cos  t  dt } + \frac{1}{2}\int t \text{ cos  3t  dt }\]
\[ = \frac{3}{2}\left[ t\int \text{ cos t dt } - \int\left( \frac{d t}{d t}\int\text{ cos  t  dt } \right)dt \right] + \frac{1}{2}\left[ t\int \text{ cos  3t  dt }- \int\left( \frac{d t}{d t}\int\text{ cos 3t  dt } \right)dt \right]\]
\[ = \frac{3}{2}\left[ t \text{ sin  t }- \int\text{ sin  t  dt } \right] + \frac{1}{2}\left[ \frac{t \sin3t}{3} - \frac{1}{3}\int\text{ sin  3t  dt } \right]\]
\[ = \frac{3}{2}\left[ t \sin t + \cos t \right] + \frac{1}{2}\left[ \frac{t \sin3t}{3} + \frac{1}{9}\cos 3t \right] + C\]
\[ = \frac{3}{2}t \sin t + \frac{3}{2}\cos t + \frac{1}{6}t \sin3t + \frac{1}{18}\cos3t + C\]
\[ = \frac{3}{2}\sqrt{x}\sin\sqrt{x} + \frac{3}{2}\cos\sqrt{x} + \frac{1}{6}\sqrt{x}\sin\left( 3\sqrt{x} \right) + \frac{1}{18}\cos\left( 3\sqrt{x} \right) + C\]

Note: The final answer in indefinite integration may vary based on the integration constant.