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. $\frac{g\left(x\right)}{g\left(\pi \right)}$

Answer 1

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

Let $I=\int {\cos }^{4}xdx$
Using $\int {\cos }^{m}x=\frac{\sin x{\cos }^{m-1}x}{m}+\frac{m-1}{m}\int {\cos }^{n-2}xdx$

$I=\frac{1}{4}\sin x{\cos }^{3}x+\frac{3}{4}\int {\cos }^{2}xdx$

$=\frac{1}{4}\sin x{\cos }^{3}x+\frac{3}{8}\int \cos 2xdx+\frac{3}{8}\int dx$

$=\frac{3x}{8}+\frac{1}{4}\sin x{\cos }^{3}x+\frac{3}{8}\sin x\cos x$

$=\frac{1}{32}(12x+8\sin 2x+\sin 4x)$
Now $g(\pi +x)=\frac{1}{32}(12(\pi +x)+8\sin \left(2(\pi +x)\right)+\sin \left(4(\pi +x)\right))$

$=\frac{1}{32}(12\pi +12x+8\sin 2x+\sin 4x)$

$=\frac{1}{32}(12x+8\sin 2x+\sin 4x)+\frac{1}{32}(12\pi +8\sin 2\pi +\sin 4\pi )=g\left(x\right)+g\left(\pi \right)$