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

1 Answer
Jan 18, 2016

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

Explanation:

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.