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) $\frac{g(x)}{g(\pi)}$
Show Answer
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 $\quad t=\pi+y$
$\Rightarrow \quad d t=d y$
$$ I _2=\int _0^{x} \cos ^{4}(y+\pi) d y $$
$$ =\int _0^{x}(-\cos y)^{4} d y=\int _0^{x} \cos ^{4} y d y=g(x) $$
$\therefore \quad g(x+\pi)=g(\pi)+g(x)$
