The integral \[ \int_{1/4}^{3/4} \cos\left( 2 \cot^{-1} \sqrt{\frac{1 - x}{1 + x}} \right) \, dx \] is equal to:
- \( -\frac{1}{2} \)
- \( \frac{1}{4} \)
- \( \frac{1}{2} \)
- \( -\frac{1}{4} \)
The Correct Option is D
Approach Solution - 1
We start with the given integral:
\[ \int_{1/4}^{3/4} \cos \left( 2 \cot^{-1} \sqrt{\frac{1 - x}{1 + x}} \right) dx \]
Let:
\[ \cot^{-1} \sqrt{\frac{1 - x}{1 + x}} = \theta \]
Then,
\[ \cot \theta = \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \Rightarrow \cos \theta = \sqrt{\frac{1 - x}{2}} \]
Now, using the double angle identity for cosine:
\[ \cos(2\theta) = 2\cos^2\theta - 1 \]
Thus, the integral becomes:
\[ \int_{1/4}^{3/4} \cos(2\theta) \, dx = \int_{1/4}^{3/4} (2\cos^2\theta - 1) \, dx \]
Substitute \(\cos^2 \theta = \frac{1 - x}{2}\):
\[ \int_{1/4}^{3/4} \left[ 2\left( \frac{1 - x}{2} \right) - 1 \right] dx = \int_{1/4}^{3/4} [(1 - x) - 1] \, dx = \int_{1/4}^{3/4} (-x) \, dx \]
Now integrate:
\[ \int_{1/4}^{3/4} (-x) \, dx = -\left[ \frac{x^2}{2} \right]_{1/4}^{3/4} \]
Substitute the limits:
\[ = -\frac{1}{2} \left[ \left( \frac{9}{16} \right) - \left( \frac{1}{16} \right) \right] = -\frac{1}{2} \times \frac{8}{16} = -\frac{1}{2} \times \frac{1}{2} = -\frac{1}{4} \]
Therefore, the value of the given integral is:
\[ \boxed{\frac{-1}{4}} \]
Approach Solution -2
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:
\[ -\frac{1}{4} \]
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