Question #ae290

Question

861 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-04T01:05:52

#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#