Question

If $g\left(x\right)={\int }_0^x{\cos }^{4}tdt.$ then $g(x+\pi )$ equals

• A. $g\left(x\right)+g\left(\pi \right)$
• B. $g\left(x\right)-g\left(\pi \right)$
• C. $g\left(x\right)g\left(\pi \right)$
• D. $\frac{g\left(x\right)}{9\left(\pi \right)}$

Answer 1

The correct option is A $g\left(x\right)+g\left(\pi \right)$

$g\left(x\right)={\int }_0^x{\cos }^{4}tdt\Rightarrow g(x+\pi )={\int }_0^{\pi +x}{\cos }^{4}tdt={\int }_0^{\pi }{\cos }^{4}tdt+{\int }_{\pi }^{\pi +x}{\cos }^{4}tdt=I_1+I_2$
Where $I_1={\int }_0^{\pi }{\cos }^{4}tdt=g\left(\pi \right)$
And $I_2={\int }_{\pi }^{\pi +x}{\cos }^{4}tdt$
Substitute $t=\pi +y\Rightarrow dt=dy$
$I_2={\int }_0^x{\cos }^4(\pi +y)dy={\int }_0^x{\cos }^{4}ydy=g\left(x\right)$
Hence $I=g(x+\pi )=g\left(\pi \right)+g\left(x\right)$