Curve integral conversion: Why use y=sin(t) instead of polar coordinate transformation?

Discussion on the replacement of curve integral variables

This paper analyzes the techniques of variable substitution in a curve integral problem. The answer is not to use polar coordinate transformation, but to use trigonometric function substitution to simplify integral calculation.

The original integral formula is: $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$

The answer is the following method of exchanging the element: Let $y = \sin(t)$. Since the integral interval $y \in (0, 1)$, then $t \in (0, \frac{\pi}{2})$. In this interval, $\sin(t)$ and $\cos(t)$ are both positive values.

After substituting the converter, the integral formula becomes:

$\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy = \int_0^{\frac{\pi}{2}} \frac{\sin^2t}{\sqrt{1-\sin^2t}}d(\sin t)$

Since $d(\sin t) = \cos t dt$, and $\sqrt{1 - \sin^2t} = \cos t$ (in the interval of $t \in (0, \frac{\pi}{2})$), the integral can be simplified to:

$\int_0^{\frac{\pi}{2}} \frac{\sin^2t}{\cos t}\cos t dt = \int_0^{\frac{\pi}{2}} \sin^2t dt$

Through the substitution of $y = \sin(t)$, the root number is cleverly eliminated and the integral calculation is simplified. This reflects the importance of choosing the appropriate variable substitution in simplifying the integration process. Compared with polar coordinate transformation, this method solves this specific integral problem more directly and effectively. The key is to reasonably select the transmutation variable and correctly handle the integral limit and differential elements.

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