What is f(x) = int sec^2x- cosx dx if f((5pi)/4) = 0 ?

1 Answer
Mar 5, 2017

f(x) = tanx-sinx+(2+sqrt2)/2

Explanation:

intsec^2x - cosxdx = intsec^2xdx-intcosxdx

These are known integrals:

d/dxtanx = sec^2x -> intsec^2xdx = tanx

d/dxsinx = cosx -> intcosxdx = sinx

therefore,

intsec^2x-cosxdx = tanx - sinx + C = f(x)

where C is the constant of integration.

The question says that f((5pi)/4)=0, so

tan((5pi)/4)-sin((5pi)/4) + C = 0

C = 1 + sqrt2/2 = (2+sqrt2)/2

Therefore,

f(x) = tanx-sinx+(2+sqrt2)/2