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If $g(x)=\int_0^x \cos ^4 t \mathrm{~d} t$, then $g(x+\pi)$ equals
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Verified Answer
The correct answer is:
$g(x)+g(\pi)$
$\begin{aligned}
& g(x)=\int_0^x \cos ^4 t \mathrm{~d} t \\
& \Rightarrow g(x+\pi)=\int_0^{x+\pi} \cos ^4 t \mathrm{~d} t=\int_0^x \cos ^4 t \mathrm{~d} t+\int_x^{x+\pi} \cos ^4 t \mathrm{~d} t \\
& =g(x)+\int_x^{x+\pi} \cos ^4 t \mathrm{~d} t \\
& =g(x)+\int_0^\pi \cos ^4 t \mathrm{~d} t
\end{aligned}$
[as $\cos ^4 t$ is a periodic function with period $\left.\pi\right]=g(x)+g(\pi)$
& g(x)=\int_0^x \cos ^4 t \mathrm{~d} t \\
& \Rightarrow g(x+\pi)=\int_0^{x+\pi} \cos ^4 t \mathrm{~d} t=\int_0^x \cos ^4 t \mathrm{~d} t+\int_x^{x+\pi} \cos ^4 t \mathrm{~d} t \\
& =g(x)+\int_x^{x+\pi} \cos ^4 t \mathrm{~d} t \\
& =g(x)+\int_0^\pi \cos ^4 t \mathrm{~d} t
\end{aligned}$
[as $\cos ^4 t$ is a periodic function with period $\left.\pi\right]=g(x)+g(\pi)$
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