Curve integral variable replacement: How to convert $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$ to $\int_0^{\frac{\pi}{2}}\sin^2tdt$?

Detailed explanation of curve integral variable replacement: Simplify the qualitative integral

This article explains in detail how to simplify the fixed integral $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$ to $\int_0^{\frac{\pi}{2}}\sin^2tdt$ by variable replacement. Many students will encounter difficulties in dealing with such points.

It is not a polar coordinate transformation, but a simple variable replacement method can be solved. The key is to choose the right replacement variable.

Problem solving steps:

We choose to replace the variable $y = \sin(t)$. Since the original integral interval is $0 \le y \le 1$, the corresponding $t$ interval is $0 \le t \le \frac{\pi}{2}$. In this interval, both $\sin(t)$ and $\cos(t)$ are non-negative.

1. Replace variable: Substitute $y = \sin(t)$ into the original integral formula: $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$

2. Replacement points limit: When $y = 0$, $t = 0$; when $y = 1$, $t = \frac{\pi}{2}$.

3. Calculate the differential: Derivative for $y = \sin(t)$ to obtain $dy = \cos(t)dt$

4. Substitute the integral formula: Substitute $y = \sin(t)$ and $dy = \cos(t)dt$ into the original integral formula:

   $\int_0^{\frac{\pi}{2}} \frac{\sin^2t}{\sqrt{1-\sin^2t}} \cos(t) dt$

5. Simplification: Since $1 - \sin^2t = \cos^2t$ and in the interval $0 \le t \le \frac{\pi}{2}$$$, $\cos(t) \ge 0$, so $\sqrt{1-\sin^2t} = \sqrt{\cos^2t} = \cos(t)$. After substituting, the integral formula becomes:

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

6. Final result: After simplification, we get:

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

Through this clever variable replacement, we successfully simplify a complex integral into a more easily computed form. Remember that choosing the right replacement variable and correctly handling the integral limit and differential terms are the key to solving the problem.

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