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

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Answer 1 · 2017-03-05T14:53:15

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

#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#