A) \[\frac{1}{32}\left[ -\frac{3}{2}\cos 2x+\frac{1}{6}\cos 6x \right]+c\]
B) \[\frac{1}{32}\left[ -\frac{3}{2}\sin 2x+\frac{1}{6}\sin 6x \right]+c\]
C) \[-\frac{1}{32}\left[ -\frac{3}{2}\cos 2x+\frac{1}{6}\sin 6x \right]+c\]
D) None of the above
Correct Answer: A
Solution :
\[\int{{{\sin }^{3}}x{{\cos }^{3}}x}dx\] \[=\frac{1}{8}\int{{{(2\sin x\cos x)}^{3}}}dx\] \[=\frac{1}{8}\int{{{\sin }^{3}}2x}dx\] \[=\frac{1}{8}\int{\left[ \frac{3\sin 2x-\sin 6x}{4} \right]}dx\] \[[\because \sin 3x=3\sin x-4{{\sin }^{3}}x\] \[\Rightarrow \]\[{{\sin }^{3}}x=\frac{3\sin x-\sin 3x}{4}]\] \[=\frac{1}{32}[3\int{\sin 2xdx-\int{\sin 6x\,dx}}]\] \[=\frac{1}{32}\left[ -\frac{3}{2}\cos 2x+\frac{1}{6}\cos 6x \right]+c\]
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