Concept:
If
f(x) is a periodic function with period T , then
∫0nTf(x)dx=n∫0Tf(x)dx If
f(x) is periodic with period
T, then
f(ax) is periodic with period
∣a∣T
Calculation:
Let f(x)=(sin4x+cos4x)⇒f(x)=(sin2x)2+(cos2x)2+2sin2xcos2x−2sin2xcos2x⇒f(x)=(sin2x+cos2x)2−2sin2xcos2x⇒f(x)=1−2sin22x⇒f(x)=1−41−cos4x⇒f(x)=43+4cos4x
Since
cosx is a function of period
2π⇒cos4x is periodic with period
42π=2π⇒f(x)=(sin4x+cos4x) is periodic with period
2π
Given,
∫0π/2(sin4x+cos4x)dx=k
So,
∫020π(sin4x+cos4x)dx=∫040(2π)(sin4x+cos4x)dx
As
f(x)=(sin4x+cos4x) is periodic with period
2π
By the property of definite integral,
∫040(2π)(sin4x+cos4x)dx=40∫0π/2(sin4x+cos4x)dx⇒∫020π(sin4x+cos4x)dx=40k{ Using (i)}