Question

If $g\left(x\right)=\sin \left[{\pi }^3\right]x+\sin [-{\pi }^3]x,$ where $[.]$ denotes the greatest integer function, then ${\int }_0^{\frac{\pi }{2}}\pi {\sin }^{4}x.g(-\frac{\pi }{2})dx$ is equal to

• A. $\frac{{\pi }^2}{16}$
• B. $\frac{{\pi }^2}{8}$
• C. $\frac{3{\pi }^2}{16}$
• D. $\frac{3{\pi }^2}{32}$

Answer 1

Answer: (C)

$\frac{3{\pi }^2}{16}$

$g\left(x\right)=\sin \left[{\pi }^3\right]x+\sin [-{\pi }^3]x=\sin \left[31.0437\right]x+\sin [-31.0437]x\Rightarrow g\left(x\right)=\sin 31x-\sin 32x$

$\therefore g(-\frac{\pi }{2})=\sin (-\frac{31}{2})\pi -\sin (-16\pi )=-\sin \frac{31}{2}\pi -0\because \sin n\pi =0\forall n\in I$

$=-\sin (\frac{\pi }{2}+15\pi )=-\cos 15\pi =-(-1{)}^{15}=1$

Now,

$\pi {\int }_0^{\frac{\pi }{2}}{\sin }^{4}x\cdot g(-\frac{\pi }{2})dx=\pi {\int }_0^{\frac{\pi }{2}}{\sin }^{4}xdx=\pi \cdot \frac{3}{4}\frac{1}{2}\frac{\pi }{2}=\frac{3{\pi }^2}{16}$