The value of `(int_(0)^(1)(dt)/(sqrt(1-t^(4))))/(int_(0)^(1)(1)/(sqrt(1+t^(4)))dt)` is

Options

A. 1
B. 2
C. `2sqrt3`
D. `sqrt2`

Explanation

To solve the problem, we need to evaluate the expression:

\[ I = \frac{\int_{0}^{1} \frac{dt}{\sqrt{1 - t^4}}}{\int_{0}^{1} \frac{dt}{\sqrt{1 + t^4}}} \]

Let's denote the numerator as \( I_1 \) and the denominator as \( I_2 \):

1. **Calculate \( I_1 \)**:
\[ I_1 = \int_{0}^{1} \frac{dt}{\sqrt{1 - t^4}} \]
We will use the substitution \( t = \sin^{1/2}(\theta) \). Then, \( dt = \frac{1}{2} \sin^{-1/2}(\theta) \cos(\theta) d\theta \) and the limits change from \( t=0 \) to \( t=1 \), giving \( \theta=0 \) to \( \theta=\frac{\pi}{2} \).

Substituting these into the integral, we get:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{2} \sin^{-1/2}(\theta) \cos(\theta)}{\sqrt{1 - \sin^2(\theta)}} d\theta \]
Using \( \sqrt{1 - \sin^2(\theta)} = \cos(\theta) \):
\[ I_1 = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin^{-1/2}(\theta)}{\cos(\theta)} d\theta \]
This simplifies to:
\[ I_1 = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{\sin(\theta)}} \]

2. **Calculate \( I_2 \)**:
\[ I_2 = \int_{0}^{1} \frac{dt}{\sqrt{1 + t^4}} \]
We will use the substitution \( t = \tan^{1/2}(\alpha) \). Then, \( dt = \frac{1}{2} \tan^{-1/2}(\alpha) \sec^2(\alpha) d\alpha \) and the limits change from \( t=0 \) to \( t=1 \), giving \( \alpha=0 \) to \( \alpha=\frac{\pi}{4} \).

Substituting these into the integral, we get:
\[ I_2 = \int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{2} \tan^{-1/2}(\alpha) \sec^2(\alpha)}{\sqrt{1 + \tan^2(\alpha)}} d\alpha \]
Using \( \sqrt{1 + \tan^2(\alpha)} = \sec(\alpha) \):
\[ I_2 = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{\tan^{-1/2}(\alpha)}{\sec(\alpha)} d\alpha \]
This simplifies to:
\[ I_2 = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{d\alpha}{\sqrt{\tan(\alpha)}} \]

3. **Combine \( I_1 \) and \( I_2 \)**:
Now we can write:
\[ I = \frac{I_1}{I_2} = \frac{\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{\sin(\theta)}}}{\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{d\alpha}{\sqrt{\tan(\alpha)}}} \]
The \( \frac{1}{2} \) cancels out:
\[ I = \frac{\int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sqrt{\sin(\theta)}}}{\int_{0}^{\frac{\pi}{4}} \frac{d\alpha}{\sqrt{\tan(\alpha)}}} \]

4. **Final Result**:
Evaluating the integrals gives:
\[ I = \sqrt{2} \]

Thus, the value of the given expression is:

\[ \boxed{\sqrt{2}} \]