What is #f(x) = int 2sin4x+secx dx# if #f(pi/4)=2 #?

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Answer 1 · 2016-01-18T13:00:43

#f(x)=-2/4cos4x+ln|secx+tanx|+0.6186#.

We may start by using standard rules of integration to find that

#f(x)=int(2sin4x+secx)dx=2intsin4xdx+intsecxdx#

#=-2/4cos4x+ln|secx+tanx|+C#.

We may now substitute the initial boundary condition into this to find the value of the constant of integration C :

#f(pi/4)=2=>-1/2cospi+ln|sec(pi/4)+tan(pi/4)|+C=2#

#therefore 1/2+ln(sqrt2+1)+C=2#

#therefore C=0.6186#.

Thus #f(x)=-2/4cos4x+ln|secx+tanx|+0.6186#.