SATHEE: Indefinite Integration 1 Question 32

Indefinite Integration 1 Question 32

33. If $g (x) = \int_{0}^{x} \cos^{4} t d t$ , then $g (x + π)$ equals

(1997, 2M)

(a) $g (x) + g (π)$

(b) $g (x) - g (π)$

(d) $\frac{g (x)}{g (π)}$

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Solution:

Given, $g (x) = \int_{0}^{x} \cos^{4} t d t$

$\Rightarrow g (x + π) = \int_{0}^{π + x} \cos^{4} t d t$

$= \int_{0}^{π} \cos^{4} t d t + \int_{π}^{π + x} \cos^{4} t d t = I_{1} + I_{2}$

where, $I_{1} = \int_{0}^{π} \cos^{4} t d t = g (π)$

and $I_{2} = \int_{π}^{π + x} \cos^{4} t d t$

Put $t = π + y$

$\Rightarrow d t = d y$

$I_{2} = \int_{0}^{x} \cos^{4} (y + π) d y$

$= \int_{0}^{x} (- \cos y)^{4} d y = \int_{0}^{x} \cos^{4} y d y = g (x)$

$∴ g (x + π) = g (π) + g (x)$

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Indefinite Integration 1 Question 32

33. If g(x)=∫0xcos4⁡tdt, then g(x+π) equals

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33. If $g (x) = \int_{0}^{x} \cos^{4} t d t$ , then $g (x + π)$ equals