Question

$ \int_{0}^{\pi /2}{x{{\sin }^{2}}x{{\cos }^{2}}x}dx $ is equal to

• A. $ \frac{{{\pi }^{2}}}{32} $
• B. $ \frac{{{\pi }^{2}}}{16} $
• C. $ \frac{\pi }{32} $
• D. none of these

Answer 1

Answer: (A)

Let $ I=\int_{0}^{\pi /2}{x{{\sin }^{2}}x{{\cos }^{2}}x}dx $ ...(i) $ I=\int_{0}^{\pi /2}{\left( \frac{\pi }{2}-x \right)}{{\sin }^{2}}x{{\cos }^{2}}x\,dx $ ...(ii) Adding Eqs. (i) and (ii) we get $ 2I=\frac{\pi }{2}\int_{0}^{\pi /2}{{{\sin }^{2}}x}{{\cos }^{2}}x\,dx $ $ =\frac{\pi }{8}\int_{0}^{\pi /2}{{{\sin }^{2}}2x}\,dx $ $ \Rightarrow $ $ 2I=\frac{\pi }{8}\left[ x-\frac{\sin 4x}{4} \right]_{0}^{\pi /2}=\frac{\pi }{8}\left[ \frac{\pi }{2} \right] $ $ \Rightarrow $ $ I=\frac{{{\pi }^{2}}}{32} $