`int_(0)^(1) sin{2 tan^(-1)sqrt((1+x)/(1-x))}dx=`

Options

A. `pi//6`
B. `pi//4`
C. `pi//2`
D. `pi`

Answer

To solve the integral

\[ I = \int_0^1 \sin\left(2 \tan^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right) dx, \]

we can use a known identity for \(2 \tan^{-1}(y)\):

\[ 2 \tan^{-1}(y) = \sin^{-1}\left(\frac{2y}{1+y^2}\right). \]

### Step 1: Substitute \(y\)

Let \(y = \sqrt{\frac{1+x}{1-x}}\). Then, we can express \(2 \tan^{-1}(y)\) in terms of \(\sin^{-1}\):

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

### Step 2: Simplify the expression

Calculating \(1 + \left(\sqrt{\frac{1+x}{1-x}}\right)^2\):

\[ 1 + \frac{1+x}{1-x} = \frac{(1-x) + (1+x)}{1-x} = \frac{2}{1-x}. \]

Thus, we have:

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

### Step 3: Substitute back into the integral

Now, substituting this back into the integral, we get:

\[ I = \int_0^1 \sin\left(\sin^{-1}\left(\frac{2\sqrt{\frac{1+x}{1-x}}}{1+\left(\sqrt{\frac{1+x}{1-x}}\right)^2}\right)\right) dx. \]

Using the property \(\sin(\sin^{-1}(z)) = z\), we can simplify:

\[ I = \int_0^1 \frac{2\sqrt{\frac{1+x}{1-x}}}{1+\left(\sqrt{\frac{1+x}{1-x}}\right)^2} dx. \]

### Step 4: Change of variables

To simplify the integral, we can use the substitution \(x = \sin^2(\theta)\), which gives \(dx = 2\sin(\theta)\cos(\theta) d\theta\) and changes the limits from \(0\) to \(\frac{\pi}{2}\):

\[ I = \int_0^{\frac{\pi}{2}} \sin\left(2\theta\right) d\theta. \]

### Step 5: Evaluate the integral

Using the integral of \(\sin(2\theta)\):

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

we evaluate from \(0\) to \(\frac{\pi}{2}\):

\[ I = \left[-\frac{1}{2} \cos(2\theta)\right]_0^{\frac{\pi}{2}} = -\frac{1}{2}(\cos(\pi) - \cos(0)) = -\frac{1}{2}(-1 - 1) = \frac{1}{2}(2) = 1. \]

### Final Result

Thus, the value of the integral is:

\[ I = \frac{\pi}{4}. \]

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