If `int1/((1+x^(2))sqrt(1-x^(2)))dx=F(x)+c " and " F(1)=0`,
then for `x gt 0, F(x)=`

Options

A. `1/sqrt(2)tan^(-1)((sqrt(2)x)/sqrt(1+x^(2)))+pi/2`
B. `1/sqrt(2)tan^(-1)((sqrt(2)x)/sqrt(1+x^(2)))-pi/(2sqrt(2))`
C. `1/sqrt(2)tan^(-1)((sqrt(2)x)/sqrt(1-x^(2)))+pi/(2sqrt(2))`
D. `1/sqrt(2)tan^(-1)((sqrt(2)x)/sqrt(1-x^(2)))-pi/(2sqrt(2))`

Answer

To solve the integral

\[ \int \frac{1}{(1+x^2)\sqrt{1-x^2}} \, dx = F(x) + C \]

with the condition \( F(1) = 0 \), we will follow these steps:

### Step 1: Substitution
Let \( x = \sin \theta \). Then, \( dx = \cos \theta \, d\theta \). The integral becomes:

\[ \int \frac{1}{(1+\sin^2 \theta)\sqrt{1-\sin^2 \theta}} \cos \theta \, d\theta \]

### Step 2: Simplifying the Integral
Using \( \sqrt{1 - \sin^2 \theta} = \cos \theta \), we rewrite the integral as:

\[ \int \frac{\cos \theta}{(1+\sin^2 \theta)\cos \theta} \, d\theta = \int \frac{1}{1+\sin^2 \theta} \, d\theta \]

### Step 3: Further Simplification
Now, we can express \( \sin^2 \theta \) in terms of \( \tan \theta \):

Let \( \tan \theta = t \). Then, \( \sin^2 \theta = \frac{t^2}{1+t^2} \) and \( d\theta = \frac{1}{1+t^2} \, dt \).

The integral becomes:

\[ \int \frac{1}{1+\frac{t^2}{1+t^2}} \cdot \frac{1}{1+t^2} \, dt = \int \frac{1+t^2}{2+t^2} \cdot \frac{1}{1+t^2} \, dt = \int \frac{1}{2+t^2} \, dt \]

### Step 4: Integrating
The integral \( \int \frac{1}{2+t^2} \, dt \) can be solved using the formula:

\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]

In our case, \( a^2 = 2 \) (so \( a = \sqrt{2} \)). Thus, we have:

\[ \int \frac{1}{2+t^2} \, dt = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \]

### Step 5: Back Substitution
Recall that \( t = \tan \theta \) and \( \theta = \sin^{-1}(x) \). Therefore, we have:

\[ F(x) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\tan(\sin^{-1}(x))}{\sqrt{2}} \right) + C \]

### Step 6: Simplifying Further
Using \( \tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1-x^2}} \), we get:

\[ F(x) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2(1-x^2)}} \right) + C \]

### Step 7: Applying the Condition \( F(1) = 0 \)
Substituting \( x = 1 \):

\[ F(1) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{1}{\sqrt{2(1-1^2)}} \right) + C = 0 \]

Since \( \tan^{-1}(\infty) = \frac{\pi}{2} \):

\[ \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2} + C = 0 \implies C = -\frac{\pi}{2\sqrt{2}} \]

### Final Result
Thus, the function \( F(x) \) is given by:

\[ F(x) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2(1-x^2)}} \right) - \frac{\pi}{2\sqrt{2}} \]