Question

If $g(x)={\int }_0^x{\cos }^{4}tdt,$ then $g(x+\pi )$ is equal to

• A. $g(x)+g(\pi )$
• B. $g(x)-g(\pi )$
• C. $g(x).g(\pi )$
• D. $\frac{g(x)}{g(\pi )}$
• E. $\frac{g(\pi )}{g(x)}$

Answer 1

Answer: (A)

$\because$ $g(x)={\int }_0^x{\cos }^{4}tdt$
$\therefore$ $g(x+\pi )={\int }_0^{\pi +x}{\cos }^{4}tdt$
$={\int }_0^{\pi }{\cos }^{4}tdt+{\int }_{\pi }^{\pi +x}{\cos }^{4}tdt$
$=I_1+I_2$
$\Rightarrow$ $I_1=g(\pi )$ and $I_2={\int }_{\pi }^{\pi +x}{\cos }^{4}tdt$
$\Rightarrow$ $I_2={\int }_0^x{\cos }^4(y+\pi )dy$
$={\int }_0^x{[\cos (\pi +y)]}^{4}dy$
$={\int }_0^x{\cos }^{4}ydy=g(x)$