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Question: \(g\left( x \right) = \int_0^x {{{\cos }^4}tdt} \), then\(g\left( {x + \pi } \right)\)equals (A) \...

g(x)=∫0xcos⁡4tdtg\left( x \right) = \int_0^x {{{\cos }^4}tdt} , theng(x+π)g\left( {x + \pi } \right)equals
(A) g(x)+g(π)g\left( x \right) + g\left( \pi \right)
(B) g(x)−g(π)g\left( x \right) - g\left( \pi \right)
(C) f(x)g(π)f\left( x \right)g\left( \pi \right)
(D) g(x)/g(π)g\left( x \right)/g\left( \pi \right)

Explanation

Solution

Hint : We have ∫acf(x)dx=∫abf(x)dx+∫ccf(x)dx\int_a^c {f\left( x \right)dx = \int_a^b {f\left( x \right)dx + \int_c^c {f\left( x \right)dx} } } we use this property of definite integral and try to get the required in terms of given options.

Complete step-by-step answer :
g(x)=∫0xcos⁡4t dtg\left( x \right) = \int_0^x {{{\cos }^4}t\,dt}
g(x+π)=∫0x+πcos⁡4t dtg\left( {x + \pi } \right) = \int_0^{x + \pi } {{{\cos }^4}t\,dt}
=∫0πcos⁡4t dt+∫πx+πcos⁡4t dt= \int_0^\pi {{{\cos }^4}t\,dt + \int_\pi ^{x + \pi } {{{\cos }^4}t\,dt} }
Now let I1=∫πx+πcos⁡4t dt{I_1} = \int_\pi ^{x + \pi } {{{\cos }^4}t\,dt}
put t=y+πt = y + \pi
⇒dt=dy\Rightarrow dt = dy
If t=π⇒y=0t = \pi \Rightarrow y = 0
t=x+π⇒y=xt = x + \pi \Rightarrow y = x
I1=∫0xcos⁡4(π+y)dy{I_1} = \int_0^x {{{\cos }^4}\left( {\pi + y} \right)dy}
=∫0xcos⁡4y dy=∫0xcos⁡4t dt= \int_0^x {{{\cos }^4}y\,dy = \int_0^x {{{\cos }^4}t\,dt} }
=g(x)= g\left( x \right)
So g(x+π)=g(x)+g(π)g\left( {x + \pi } \right) = g\left( x \right) + g\left( \pi \right)
So, the correct answer is “Option A”.

Note : When we use a substitution method to solve any integral always remember that we must change the limits as well accordingly. If we do not change the limits after applying substitution methods then the final answer obtained will be wrong.