Definite Integration Question 32
Question 32
- If $g(x)=\int_{0}^{x} \cos ^{4} t d t$, then $g(x+\pi)$ equals
(1997, 2M) (a) $g(x)+g(\pi)$ (b) $g(x)-g(\pi)$ (c) $g(x) g(\pi)$ (d) $\frac{g(x)}{g(\pi)}$
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Solution:
- Given, $g(x)=\int_{0}^{x} \cos ^{4} t d t$
$\Rightarrow \quad g(x+\pi)=\int_{0}^{\pi+x} \cos ^{4} t d t$
$$ =\int_{0}^{\pi} \cos ^{4} t d t+\int_{\pi}^{\pi+x} \cos ^{4} t d t=I_{1}+I_{2} $$
where, $\quad I_{1}=\int_{0}^{\pi} \cos ^{4} t d t=g(\pi)$
and $\quad I_{2}=\int_{\pi}^{\pi+x} \cos ^{4} t d t$
Put $\quad t=\pi+y$
$\Rightarrow \quad d t=d y$
$$ I_{2}=\int_{0}^{x} \cos ^{4}(y+\pi) d y $$
$$ =\int_{0}^{x}(-\cos y)^{4} d y=\int_{0}^{x} \cos ^{4} y d y=g(x) $$
$\therefore \quad g(x+\pi)=g(\pi)+g(x)$
