Question #ea4c4

1 Answer
Nov 15, 2017

f(t)=12arctg(t)-3Pi

Explanation:

we know

f'(t)=dy/dt

dy/dt=(12)/(1+t^2)

transposing differentials

dy=(12)/(1+t^2)dt

integrating

int(12)/(1+t^2)dt

t=tan(u)
dt=1/cos^2(u)du

int(12)/((cos^2(u)/(cos^2(u))+(sin^2(u))/(cos^2(u))))(1/cos^2(u)du)

cos^2(u)+sin^2(u)=1

int(12)/((1/(cos^2(u))))(1/cos^2(u)du)

dividing

int(12)du

integrating

int(12)du=12u+c

but arctg(t)=u
then

12u+c=12arctg(t)+c

f(t)=12arctg(t)+c ... 1

now finding the constant 'c' f(1)=0

in (1)

f(1)=12arctg(1)+c

0=12arctg(1)+c

arctg(1)=Pi/4

0=12Pi/4+c

0=3Pi+c

-3Pi=c

now

f(t)=12arctg(t)-3Pi