Quick question on integral distribution

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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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