我的提问

Evaluate the integral using trigonometric substitution

\int \frac{1}{y^2 \sqrt{10 - y^2}} \, dy

o1的回答

To evaluate the integral

\int \frac{1}{y^2 \sqrt{10 - y^2}} \, dy,

we can use trigonometric substitution. Here's a step-by-step solution:

The presence of \sqrt{10 - y^2} suggests using the substitution:

y = \sqrt{10} \sin \theta

This substitution leverages the Pythagorean identity, which simplifies the square root.

Differentiate y with respect to \theta:

dy = \sqrt{10} \cos \theta \, d\theta

Substitute y and dy into the integral:

\int \frac{1}{y^2 \sqrt{10 - y^2}} \, dy = \int \frac{1}{10 \sin^2 \theta \cdot \sqrt{10 - 10 \sin^2 \theta}} \cdot \sqrt{10} \cos \theta \, d\theta

Simplify the square root:

\sqrt{10 - 10 \sin^2 \theta} = \sqrt{10 (1 - \sin^2 \theta)} = \sqrt{10} \cos \theta

Substitute back:

\int \frac{1}{10 \sin^2 \theta \cdot \sqrt{10} \cos \theta} \cdot \sqrt{10} \cos \theta \, d\theta = \frac{1}{10} \int \csc^2 \theta \, d\theta

The integral of \csc^2 \theta is straightforward:

\frac{1}{10} \int \csc^2 \theta \, d\theta = -\frac{1}{10} \cot \theta + C

From the original substitution y = \sqrt{10} \sin \theta, we can find \cot \theta:

\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{10 - y^2}/\sqrt{10}}{y/\sqrt{10}} = \frac{\sqrt{10 - y^2}}{y}

Thus, the integral in terms of y is:

-\frac{1}{10} \cdot \frac{\sqrt{10 - y^2}}{y} + C = -\frac{\sqrt{10 - y^2}}{10 y} + C

After simplifying, an antiderivative is -\sqrt{10 - y^2} divided by 10y plus C. Thus,

-\frac{\sqrt{10 - y^2}}{10 y} + C