Quick question on integral distribution

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SUMMARY

The discussion centers on the integral distribution of the expression Integral of[1/4 + cos2x/2 + 1/4((1+cos4x)/2)] dx. The confusion arises from the transformation to 3/8Integral dx + 1/4integral 2cos2xdx + 1/32integral 4cos4x dx. The value 3/8 is derived from the sum of 1/4 from the first term and 1/8 from the last term after expanding and regrouping the integrals. This clarification resolves the misunderstanding regarding the distribution of the integral.

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frasifrasi
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So one example in my book goes from this step:

Integral of[1/4 + cos2x/2 + 1/4((1+cos4x)/2)] dx

to

= 3/8Integral dx + 1/4integral 2cos2xdx + 1/32integral 4cos4x dx

I am very confused as to why, in the process od destribution, the integral became 3/8intdx as opposed to just int 1/4dx and so forth... Where did that 3/8 come form?

Thank you!
 
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Integral of[1/4 + cos2x/2 + 1/4((1+cos4x)/2)] dx

Expand the third term and regroup terms of dx, cos 2x dx, cos 4x dx
 
frasifrasi said:
So one example in my book goes from this step:

Integral of[1/4 + cos2x/2 + 1/4((1+cos4x)/2)] dx

to

= 3/8Integral dx + 1/4integral 2cos2xdx + 1/32integral 4cos4x dx

I am very confused as to why, in the process od destribution, the integral became 3/8intdx as opposed to just int 1/4dx and so forth... Where did that 3/8 come form?

Thank you!
3/8= 1/4+ 1/8. The 1/4 is the first term and the 1/8 is from the last term.
 

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