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 :
$$\eqalign{
& {\text{Given that }}\,g\left( x \right) = \int\limits_0^x {{{\cos }^4}t\,dt} \cr
& \therefore g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {{{\cos }^4}t\,dt} \cr
& = \int\limits_0^\pi {{{\cos }^4}t\,dt} + \int\limits_\pi ^{x + \pi } {{{\cos }^4}t\,dt} \cr
& g\left( {x + \pi } \right) = g\left( \pi \right) + I,{\text{ where }}I = \int\limits_\pi ^{x + \pi } {{{\cos }^4}t\,dt} \cr
& {\text{Put }}t = \pi + y,\,dt = dy \cr
& {\text{Also as }}t \to \pi ,\,y \to 0 \cr
& {\text{as }}t \to x + \pi ,\,y \to x \cr
& \therefore I = \int\limits_0^x {{{\cos }^4}\left( {\pi + y} \right)\,dy} \cr
& = \int\limits_0^x {{{\cos }^4}y\,dy} = \int\limits_0^x {{{\cos }^4}t\,dt} = g\left( x \right) \cr
& \therefore g\left( {x + \pi } \right) = g\left( \pi \right) + g\left( x \right) \cr} $$