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Curve integral variable replacement: How to convert $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$ to $\int_0^{\frac{\pi}{2}}\sin^2tdt$?
Curve integral variable replacement: How to convert $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$ to $\int_0^{\frac{\pi}{2}}\sin^2tdt$?
Apr 19, 2025 pm 03:03 PM
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.
Replace variable: Substitute $y = \sin(t)$ into the original integral formula: $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$
Replacement points limit: When $y = 0$, $t = 0$; when $y = 1$, $t = \frac{\pi}{2}$.
Calculate the differential: Derivative for $y = \sin(t)$ to obtain $dy = \cos(t)dt$
-
Substitute the integral formula: Substitute $y = \sin(t)$ and $dy = \cos(t)dt$ into the original integral formula:
$\int_0^{\frac{\pi}{2}} \frac{\sin^2t}{\sqrt{1-\sin^2t}} \cos(t) dt$
-
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:
$\int_0^{\frac{\pi}{2}} \frac{\sin^2t}{\cos(t)} \cos(t) dt$
-
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.
The above is the detailed content of Curve integral variable replacement: How to convert $\int_0^1 \frac{y^2}{\sqrt{1-y^2}}dy$ to $\int_0^{\frac{\pi}{2}}\sin^2tdt$?. For more information, please follow other related articles on the PHP Chinese website!
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