I'm struggling with a question and I can't solve it, I can't find good parameterization so I will be able to calculate the line Integral directly.
The question :
Given the line integral
$C$ : $y=x^3$ from $(0,0) \rightarrow (1,1)$.
Calculate the following integral directly (without Green's theorem) :
$\int_c (y+tan^3(x))dx+(3x-tan^3(y))dy$
My try :
Well I tried the 'default' scenario of $y=f(t)$ but it failed ..
$$x=t \quad y=t^3 \quad 0\le t \le 1$$
$$ dx=dt \quad dy=3t^2 dt$$
$\int_0^1 t^3+tan^3(t)+(3t-tan^3(t^3))3t^2$
I reached till here :
$2.5 +\int_0^1 tan^3(t)-3t^2tan^3(t^3)dt$
I solved the line integral using Green's theorem and found out he is equal to $2.5$ as written , so the integral must be equal to $0$.
Is there any better parameterization to solve it directly than the one I did ?
Thank you in advance!
Best Answer
Hint:-
$\int_{0}^{1}3t^{2}\tan^{3}(t^{3})\,dt=\int_{0}^{1}\tan^{3}(z)\,dz$ with substitution $z=t^{3}$
You directly end up with $2.5+\int_{0}^{1}\tan^{3}(t)\,dt-\int_{0}^{1}\tan^{3}(z)\,dz$.
In integration, the variable wrt which you are integrating does not matter.
$\int_{a}^{b} f(x)\,dx = \int_{a}^{b}f(\theta)\,d\theta=\int_{a}^{b}f(\phi)\,d\phi$.
It does not matter what variable you use. Ultimately, the definite integral is just a real number (If it exists).