If gx-0xcos4tdt- then gx- equals to

Given : nbsp; g(x)=∫_0^xcos^4tdt ⇒ g(x+π)=∫_0^x+πcos^4tdt =∫_0^πcos^4tdt+∫_π^x+πcos^4tdt =g(π)+I Here I=∫_π^x+πcos^4tdt Substitute t=y+π⇒ dt=dy ⇒ I=∫_0^xcos ...

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Byju's Answer

Standard XII

Mathematics

How to Find the Inverse of a Function

If gx=∫0xcos4...

Question

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

g (x) + g (π)

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g (x) - g (π)

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g (x) g (π)

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\frac{g (x)}{g (π)}

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Solution

The correct option is A $g (x) + g (π)$
Given :
$g (x) = x \int 0 {cos}^{4} t d t$
$\Rightarrow g (x + π) = x + π \int 0 {cos}^{4} t d t = π \int 0 {cos}^{4} t d t + x + π \int π {cos}^{4} t d t = g (π) + I$
Here $I = x + π \int π {cos}^{4} t d t$
Substitute $t = y + π \Rightarrow d t = d y$
$\Rightarrow I = x \int 0 {cos}^{4} (π + y) d y \Rightarrow I = x \int 0 {cos}^{4} (y) d y = g (x)$
So, $g (x + π) = g (x) + g (π)$

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If g(x)=x∫0cos4tdt, then g(x+π) equals to

The correct option is A g(x)+g(π)Given : g(x)=x∫0cos4tdt ⇒g(x+π)=x+π∫0cos4tdt=π∫0cos4tdt+x+π∫πcos4tdt=g(π)+I Here I=x+π∫πcos4tdt Substitute t=y+π⇒dt=dy ⇒I=x∫0cos4(π+y)dy⇒I=x∫0cos4(y)dy=g(x) So, g(x+π)=g(x)+g(π)

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