URGENT!!! Advanced Calculus Question? 10 points for informative answer!

Hello Experts,

How to compute the integral (with and without Green's theorem):

int ( (y+tan^3(x)) dx + (3x-tan^3(y)) dy) above C

Where C is the curve y = x^3 from (0,0) to (1,1).

Please give me a step by step answer.

Let c denote the curve y = x from (0,0) to (1,1), and let D denote the region in the xy-plane between the two curves. I'll write int_C, int_c, and int_D for the corresponding integrals.

By Green's theorem,
(int_C - int_c) ( (y+tan^3(x)) dx + (3x-tan^3(y)) dy) = int_D (2) dA

Therefore we need to compute
int_c ( (y+tan^3(x)) dx + (3x-tan^3(y)) dy) = int_c (4x dx)
= 2x^2 (from x = 0 to x = 1)
= 2

int_D (2) dA = int(2x - 2x^3, x = 0 to 1)
= x^2 - 1/2 x^4 (from x = 0 to x = 1)
= 1/2

Thus,
int_C ( (y+tan^3(x)) dx + (3x-tan^3(y)) dy)
= 2 + 1/2
= 5/2.

Here is a direct solution, without using Green's theorem. Parametrize the curve: x=t, y=t^3, 0 <= t <=1. Plug in for x, dx, y, and dy in the integral:
int(0 to 1) (t^3+tan^3(t))dt+(3t-tan^3(t^3))3t^2 dt=
=int(0 to 1) (t^3+tan^3(t)+9t^3-3t^2tan^3(t^3)) dt=
=int(0 to 1) (10t^3+tan^3(t)-3t^2 tan^3(t^3)) dt=
=5/2 t^4 + 1/2 sec^2(t)+ln|cos(t)|-1/2 sec^2(t^3)-ln|cos(t^3)| evaluated at 1 and 0 (see below)
=5/2 + 1/2 sec^2(1)+ln|cos(1)|-1/2 sec^2(1)-ln|cos(1)|-1/2+1/2=
=5/2.

To integrate tan^3(t), rewrite it as tan(t)tan^2(t)=tan(t)(sec^2(t)-1)=tan(t)… - tan(t)= sec(t)sec(t) tan(t)-tan(t). Use substitution with u=sec(t) to integrate the first term, and use substitution with u=cos(t) to integrate the second term.

Will edit with a solution using Green's theorem if I find one.

Christ i don't know, ask a teacher or something