Question

Sketch the graph and find the area under the curve defined by the parametric equations x= tan^-1t
y= t^3, for 0<=t<=1

Answer 1

since tan(x) = t

y=(tan(x))^3

Answer 2

when t=0, x=0

when t=1, x=pi/4

Answer 3

tan^2 tan

tan^2 = sec^2-1

sec^2(tan) - tan

sec(sectan) ints up to 1/2 sec^2

-tan ints up to ln|cos| i believe

Answer 4

so; we should have:

1/2 sec^2(x) + ln|cos(x)| from 0 to pi/4 as a solution if i see it right

Answer 5

you lost me here sec(sectan) ints up to 1/2 sec^2

Answer 6

i wonder if thats the same as integrating r

r = <tan-1(t), t^3>

R = < ?? ,t^4/4> from 0 to 1

I cant recall in itegral for tan-1 at the moment :)

Answer 7

yeah

1/2 sec^2 derives to:

2/2 sec * sec tan

Answer 8

therefore, sec sectan ints up to 1/2 sec^2

Answer 9

oh ok you took the integral

Answer 10

yep :)