Question #ae290

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Answer 1 · 2017-12-04T01:05:52

$f (x) = tan x - cot x - x + 1 + \frac{π}{4}$ $f(x)=tanx-cotx-x+1+pi/4$

$f'(x)=sec^2x+cosec^2x-1$

integrating

$intf'(x)dx=intsec^2x+cosec^2x-1dx$

$f(x)=intsec^2xdx+intcosec^2xdx-int1dx$

$intsec^2xdx=tanx+c$

$intcosec^2xdx=-cotx+c$

$int1dx=x+c$

$f(x)=intsec^2xdx+intcosec^2xdx-int1dx=tanx-cotx-x+c$

$f(x)=tanx-cotx-x+c$

finding the constant

when $f(pi/4)=1$

$f(pi/4)=tan(pi/4)-cot(pi/4)-(pi/4)+c=1$

$f(pi/4)=1-1-(pi/4)+c=1$

$-(pi/4)+c=1$

$c=1+pi/4$

now f(x) is

$f(x)=tanx-cotx-x+1+pi/4$

Answer 2 · 2017-12-04T01:06:02

$f(x)=tanx-cotx-x+pi/4+1$

.

To find $f(x)$ from $f'(x)$ , we have to take the integral of $f'(x)$ .

$int(sec^2x+csc^2x-1)dx =intsec^2dx+intcsc^2xdx-intdx$

$f(x)=tanx-cotx-x+C$

Now, we can plug in our initial conditions to solve for C:

$f(pi/4)=tan(pi/4)-cot(pi/4)-pi/4+C=1$

$1-1-pi/4+C=1$

$cancelcolor(red)1-cancelcolor(red)1-pi/4+C=1$

$-pi/4+C=1$

$C=1+pi/4$

$f(x)=tanx-cotx-x+pi/4+1$