What is $f (x) = \int {sec}^{2} x - cos x d x$ $f(x) = int sec^2x- cosx dx$ if $f (\frac{5 π}{4}) = 0$ $f((5pi)/4) = 0$ ?

Question

1437 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-11T18:54:19

$f (x) = tan x - sin x - (1 + \frac{\sqrt{2}}{2})$ $f(x)=tanx-sinx-(1+sqrt2/2)$

$\int ({sec}^{2} x - cos x) d x$ $int(sec^2x-cosx)dx$

$f (x) = tan x - sin x + c$ $f(x)=tanx-sinx+c$

$f (\frac{5 π}{4}) = tan (\frac{5 π}{4}) - sin (\frac{5 π}{4}) + c = 0$ $f((5pi)/4)=tan((5pi)/4)-sin((5pi)/4)+c=0$

$1 - (- \frac{\sqrt{2}}{2}) + c = 0$ $1-(-sqrt2/2)+c=0$

$c = - (1 + \frac{\sqrt{2}}{2})$ $c=-(1+sqrt2/2)$

$f (x) = tan x - sin x - (1 + \frac{\sqrt{2}}{2})$ $f(x)=tanx-sinx-(1+sqrt2/2)$

What is f(x) = int sec^2x- cosx dxf(x)=∫sec2x−cosxdxf(x) = int sec^2x- cosx dx if f((5pi)/4) = 0 f(5π4)=0f((5pi)/4) = 0 ?