If ${\rm{g}}\left( x \right) = \mathop \smallint \nolimits_0^x \,\,{\cos ^4}\,\,t\,dt$, then ${\rm{g}}\left( {x\,\, + \,\,\pi } \right)$ equals

${\rm{g}}\left( x \right)\,\, + \,\,{\rm{g}}\left( \pi \right)$

${\rm{g}}\left( x \right)\,\, - \,\,{\rm{g}}\left( \pi \right)$

${\rm{g}}\left( x \right)\,\,{\rm{g}}\left( \pi \right)$

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

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

\mathrm{g}\left( x \right) = \int \nolimits_0^x \,\,{\cos ^4}\,\,t\,dt

, then

\mathrm{g}\left( {x\,\, + \,\,\pi } \right)

equals

View answer and explanation

Correct option: A

\mathrm{g}\left( x \right)\,\, + \,\,\mathrm{g}\left( \pi \right)

Option A (g(x) + g(π)) is correct because substituting t = π + y allows us to separate the integral into g(x) + g(π).