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  4. If g(x)=∫0x cos 4t dt, then g(x+π ) is equal to

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Q. If $ g(x)=\int_{0}^{x}{{{\cos }^{4}}}t\,dt, $ then $ g(x+\pi ) $ is equal to

KEAMKEAM 2007Integrals

Solution:

$ \because $ $ g(x)=\int_{0}^{x}{{{\cos }^{4}}t\,dt} $
$ \therefore $ $ g(x+\pi )=\int_{0}^{\pi +x}{{{\cos }^{4}}t}\,dt $
$=\int_{0}^{\pi }{{{\cos }^{4}}t}\,dt+\int_{\pi }^{\pi +x}{{{\cos }^{4}}t}\,dt $
$={{I}_{1}}+{{I}_{2}} $
$ \Rightarrow $ $ {{I}_{1}}=g(\pi ) $ and $ {{I}_{2}}=\int_{\pi }^{\pi +x}{{{\cos }^{4}}t}\,dt $
$ \Rightarrow $ $ {{I}_{2}}=\int_{0}^{x}{{{\cos }^{4}}(y+\pi )}\,dy $
$=\int_{0}^{x}{{{[\cos (\pi +y)]}^{4}}}dy $
$=\int_{0}^{x}{{{\cos }^{4}}y\,}dy=g(x) $