Question

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

Answer 1

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