Question
If $$g\left( x \right) = \int_0^x {{{\cos }^4}t\,dt} $$ then $$g\left( {x + \pi } \right)$$ 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 :
$$g\left( x \right) + g\left( \pi \right)$$
Solution :
$$\eqalign{
& g\left( x \right) = \int_0^x {{{\left( {\frac{{1 + \cos \,2t}}{2}} \right)}^2}dt} \cr
& = \frac{1}{4}\int_0^x {\left( {1 + 2\cos \,2t + {{\cos }^2}2t} \right)} dt \cr
& = \frac{1}{4}\left[ {t + \sin \,2t} \right]_0^x + \frac{1}{8}\int_0^x {\left( {1 + \cos \,4t} \right)dt} \cr
& = \frac{1}{4}\left( {x + \sin \,2x} \right) + \frac{1}{8}\left[ {t + \frac{{\sin \,4t}}{4}} \right]_0^x \cr
& = \frac{1}{4}\left( {x + \sin \,2x} \right) + \frac{1}{8}\left( {x + \frac{{\sin \,4x}}{4}} \right) \cr
& = \frac{3}{8}x + \frac{1}{4}\sin \,2x + \frac{1}{{32}}\sin \,4x \cr
& \therefore \,g\left( {x + \pi } \right) = \frac{3}{8}\left( {x + \pi } \right) + \frac{1}{4}\sin \,2x + \frac{1}{{32}}\sin \,4x \cr
& = \frac{3}{8}\pi + g\left( x \right) \cr
& = g\left( \pi \right) + g\left( x \right) \cr} $$