Question

if $g(x)={\int }_0^x{\cos }^{4}tdt$ , then $g(x+\pi )$ equals

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

Answer 1

Answer: (A)

Given, $g(x)=\underset{0}{\overset{x}{\int }}{\cos }^{4}tdt$
$\Rightarrow g(x+\pi )=\underset{0}{\overset{\pi +x}{\int }}{\cos }^{4}tdt$
$=\underset{0}{\overset{\pi }{\int }}{\cos }^{4}tdt+\underset{\pi }{\overset{\pi +x}{\int }}{\cos }^{4}tdt$
$=I_1+I_2$
where, $I_1=\underset{0}{\overset{\pi }{\int }}{\cos }^{4}tdt=g(\pi )$
and $I_2=\underset{\pi }{\overset{\pi +x}{\int }}{\cos }^{4}tdt$
Put $t=\pi +y$
$\Rightarrow t=dy$
$I_2=\underset{0}{\overset{x}{\int }}{\cos }^4(y+\pi )dy$
$=\underset{0}{\overset{x}{\int }}(-\cos y{)}^{4}dy=\underset{0}{\overset{x}{\int }}{\cos }^{4}ydy$
$=g(x)$
$\therefore g(x+\pi )=g(\pi )+g(x)$