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

1886 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-05T14:53:15

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

$\int {sec}^{2} x - cos x d x = \int {sec}^{2} x d x - \int cos x d x$ $intsec^2x - cosxdx = intsec^2xdx-intcosxdx$

These are known integrals:

$\frac{d}{d x} tan x = {sec}^{2} x \to \int {sec}^{2} x d x = tan x$ $d/dxtanx = sec^2x -> intsec^2xdx = tanx$

$\frac{d}{d x} sin x = cos x \to \int cos x d x = sin x$ $d/dxsinx = cosx -> intcosxdx = sinx$

therefore,

$\int {sec}^{2} x - cos x d x = tan x - sin x + C = f (x)$ $intsec^2x-cosxdx = tanx - sinx + C = f(x)$

where $C$ $C$ is the constant of integration.

The question says that $f (\frac{5 π}{4}) = 0$ $f((5pi)/4)=0$ , so

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

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

Therefore,

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

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 ?