Question

The integral $ \int{{{\sin }^{3}}x}{{\cos }^{3}}x\,dx $ is equal to

• 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

Answer 1

Answer: (A)

$ \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 $