Answer
\[ - \frac{2}{{\sqrt {\sin x} }} + - \frac{2}{3}{\sin ^{\frac{3}{2}}}x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sin }^{\frac{{ - 3}}{2}}}x{{\cos }^3}xdx} \hfill \\
\hfill \\
write\,\,{\cos ^3}x\,\,as\,\,{\cos ^2}x\cos x\,\, \hfill \\
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= \int_{}^{} {{{\sin }^{\frac{{ - 3}}{2}}}x{{\cos }^3}xdx} = \int_{}^{} {{{\sin }^{\frac{{ - 3}}{2}}}x{{\cos }^2}x\cos xdx} \hfill \\
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\,\,{\cos ^2}x = 1 - {\sin ^2}x \hfill \\
\hfill \\
= \int_{}^{} {{{\sin }^{\frac{{ - 3}}{2}}}x{{\cos }^2}x\cos xdx} \hfill \\
\hfill \\
= \int_{}^{} {{{\sin }^{\frac{{ - 3}}{2}}}x\,\left( {1 - {{\sin }^2}x} \right)\cos xdx} \hfill \\
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multiply \hfill \\
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= \int_{}^{} {\,\left( {{{\sin }^{\frac{{ - 3}}{2}}}x - {{\sin }^{\frac{1}{2}}}x} \right)\cos xdx} \hfill \\
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set\,\,\,\,u = \sin x\,\,then\,\,\,du = \cos xdx \hfill \\
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\int_{}^{} {\,\left( {{u^{\frac{{ - 3}}{2}}} - {u^{\frac{1}{2}}}} \right)du} \hfill \\
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{\text{integrate}} \hfill \\
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\frac{{{u^{\frac{{ - 1}}{2}}}}}{{ - \frac{1}{2}}} - \frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}} + C \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
- \frac{2}{{{u^{\frac{1}{2}}}}} - \frac{2}{3}{u^{\frac{3}{2}}} + C \hfill \\
\hfill \\
- \frac{2}{{\sqrt u }} - \frac{2}{3}{u^{\frac{3}{2}}} + C \hfill \\
\hfill \\
\,substitute\,\,back\,\,u = \sin x \hfill \\
\hfill \\
- \frac{2}{{\sqrt {\sin x} }} + - \frac{2}{3}{\sin ^{\frac{3}{2}}}x + C \hfill \\
\hfill \\
\end{gathered} \]
