sequence and series, Mathematics

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discuss the infinite series

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integration: if f(x)+f(x+1/2) =1 find limit 0 to 2, f(x)+f(x+1/2) =1 f(x...

f(x)+f(x+1/2) =1 f(x)=1-f(x+1/2) 0∫2f(x)dx=0∫21-f(x+1/2)dx 0∫2f(x)dx=2-0∫2f(x+1/2)dx take (x+1/2)=v dx=dv 0∫2f(v)dv=2-0∫2f(v)dv 2(0∫2f(v)dv)=2 0∫2f(v)dv=1 0∫2f(x)dx=1

integral 0 to pi e^cosx cos (sinx) dx, Let u = sin(x). Then du = cos(x) dx...

Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This is e^u + C = e^sin(x) + C.  Then substitute your range 0 to pi. e^sin (pi)-e^sin(0) =0-0 =0

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