What is $f (x) = \int 2 sin 4 x + sec x d x$ $f(x) = int 2sin4x+secx dx$ if $f (\frac{π}{4}) = 2$ $f(pi/4)=2$ ?

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What is $f (x) = \int 2 sin 4 x + sec x d x$ $f(x) = int 2sin4x+secx dx$ if $f (\frac{π}{4}) = 2$ $f(pi/4)=2$ ?

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Answer 1 · 2016-01-18T13:00:43

$f (x) = - \frac{2}{4} cos 4 x + ln | sec x + tan x | + 0.6186$ $f(x)=-2/4cos4x+ln|secx+tanx|+0.6186$ .

Explanation:

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

$f (x) = \int (2 sin 4 x + sec x) d x = 2 \int sin 4 x d x + \int sec x d x$ $f(x)=int(2sin4x+secx)dx=2intsin4xdx+intsecxdx$

$= - \frac{2}{4} cos 4 x + ln | sec x + tan x | + C$ $=-2/4cos4x+ln|secx+tanx|+C$ .

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

$f (\frac{π}{4}) = 2 \Rightarrow - \frac{1}{2} cos π + ln ∣ ∣ sec (\frac{π}{4}) + tan (\frac{π}{4}) ∣ ∣ + C = 2$ $f(pi/4)=2=>-1/2cospi+ln|sec(pi/4)+tan(pi/4)|+C=2$

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

$therefore C=0.6186$ .

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

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What is $f (x) = \int 2 sin 4 x + sec x d x$ $f(x) = int 2sin4x+secx dx$ if $f (\frac{π}{4}) = 2$ $f(pi/4)=2$ ?

1 Answer

Explanation:

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