Question

The integral \[ \int_{1/4}^{3/4} \cos\left( 2 \cot^{-1} \sqrt{\frac{1 - x}{1 + x}} \right) \, dx \] is equal to:

• A. \( -\frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( -\frac{1}{4} \)

Answer 1

The Correct Option is D

Given:

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

Therefore:

\[ \cot(\theta) = \sqrt{\frac{1 - x}{1 + x}} \implies \tan(\theta) = \sqrt{\frac{1 + x}{1 - x}} \]

Using the double-angle formula for cosine:

\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]

Now express \(\sin^2(\theta)\):

\[ \sin^2(\theta) = \frac{1}{1 + \cot^2(\theta)} \]

Substituting \(\cot^2(\theta) = \frac{1 - x}{1 + x}\), we get:

\[ \sin^2(\theta) = \frac{1}{1 + \frac{1 - x}{1 + x}} = \frac{1 + x}{2} \]

Thus:

\[ \cos(2\theta) = 1 - 2\sin^2(\theta) = 1 - 2 \cdot \frac{1 + x}{2} = -x \]

The integral simplifies to:

\[ \int_{1/4}^{3/4} \cos \left( 2 \cot^{-1} \sqrt{\frac{1 - x}{1 + x}} \right) dx = \int_{1/4}^{3/4} -x dx \]

Evaluate the simplified integral:

\[ \int_{1/4}^{3/4} -x dx = - \int_{1/4}^{3/4} x dx \]

The integral of \(x\) is:

\[ \int x dx = \frac{x^2}{2} \]

Evaluate the limits:

\[ - \left[ \frac{x^2}{2} \right]_{1/4}^{3/4} = - \left( \frac{\left(\frac{3}{4}\right)^2}{2} - \frac{\left(\frac{1}{4}\right)^2}{2} \right) \]

Simplify:

\[ - \left( \frac{9}{32} - \frac{1}{32} \right) = - \frac{8}{32} = -\frac{1}{4} \]

Final Answer:

\[ -\frac{1}{4} \]