Indefinite Integration 1 Question 32

33. 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:

  1. 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 $\quad t=\pi+y$

$\Rightarrow \quad d t=d y$

$$ I _2=\int _0^{x} \cos ^{4}(y+\pi) d y $$

$$ =\int _0^{x}(-\cos y)^{4} d y=\int _0^{x} \cos ^{4} y d y=g(x) $$

$\therefore \quad g(x+\pi)=g(\pi)+g(x)$



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