`int_0^1sin(2tan^(-1)sqrt((1+x)/(1-x)))dx` is equal to

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 will follow these steps:

### Step 1: Substitution
Let \( x = \cos \theta \). Then, \( dx = -\sin \theta \, d\theta \). The limits change as follows:
- When \( x = 0 \), \( \theta = \frac{\pi}{2} \).
- When \( x = 1 \), \( \theta = 0 \).

Thus, the integral becomes:
\[ I = \int_{\frac{\pi}{2}}^0 \sin\left(2\tan^{-1}\left(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)\right)(-\sin \theta \, d\theta) \]
This simplifies to:
\[ I = \int_0^{\frac{\pi}{2}} \sin\left(2\tan^{-1}\left(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)\right)\sin \theta \, d\theta \]

### Step 2: Simplifying the Argument of Sine
Next, we simplify \( \tan^{-1}\left(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right) \):
\[ \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} = \cot\left(\frac{\theta}{2}\right) \]
Thus, \( \tan^{-1}\left(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right) = \frac{\pi}{2} - \frac{\theta}{2} \).

### Step 3: Substitute Back
Now substituting back into the integral:
\[ I = \int_0^{\frac{\pi}{2}} \sin\left(2\left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) \sin \theta \, d\theta \]
This simplifies to:
\[ I = \int_0^{\frac{\pi}{2}} \sin\left(\pi - \theta\right) \sin \theta \, d\theta = \int_0^{\frac{\pi}{2}} \sin \theta \sin \theta \, d\theta = \int_0^{\frac{\pi}{2}} \sin^2 \theta \, d\theta \]

### Step 4: Evaluate the Integral
Using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \):
\[ I = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}} \]
Calculating the limits:
\[ = \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{4} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{\pi}{4} \]