Question

Evaluate \[ \int_{0}^{1}\cot^{-1}(1+x+x^2)\,dx \]

• A. \(2\tan^{-1}2-\frac{1}{2}\ln\left(\frac{5}{4}\right)-\frac{\pi}{2}\)
• B. \(2\tan^{-1}2-\frac{1}{2}\ln\left(\frac{5}{4}\right)+\frac{\pi}{2}\)
• C. \(2\tan^{-1}2+\frac{1}{2}\ln\left(\frac{5}{4}\right)-\frac{\pi}{2}\)
• D. \(2\tan^{-1}2+\frac{1}{2}\ln\left(\frac{5}{4}\right)+\frac{\pi}{2}\)

Hint:

Many inverse trigonometric integrals simplify using the identity \(\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left(\frac{A-B}{1+AB}\right)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The identity \( \cot^{-1} \theta = \tan^{-1}(1/\theta) \) and the property \( \tan^{-1} A - \tan^{-1} B = \tan^{-1}(\frac{A-B}{1+AB}) \) are key here.
Step 2: Key Formula or Approach:
\[ \cot^{-1}(1 + x + x^2) = \tan^{-1}\left(\frac{1}{1 + x(x+1)}\right) = \tan^{-1}\left(\frac{(x+1) - x}{1 + x(x+1)}\right) \] \[ = \tan^{-1}(x+1) - \tan^{-1} x \]
Step 3: Detailed Explanation:
1. The integral becomes: \( I = \int_0^1 \tan^{-1}(x+1) \, dx - \int_0^1 \tan^{-1} x \, dx \). 2. Use integration by parts \( \int \tan^{-1} u \, du = u \tan^{-1} u - \frac{1}{2} \ln(1 + u^2) \). 3. First part: \( [ (x+1)\tan^{-1}(x+1) - \frac{1}{2}\ln(1+(x+1)^2) ]_0^1 \) \[ = (2\tan^{-1} 2 - \frac{1}{2}\ln 5) - (1\tan^{-1} 1 - \frac{1}{2}\ln 2) \] 4. Second part: \( [ x\tan^{-1} x - \frac{1}{2}\ln(1+x^2) ]_0^1 = (1\tan^{-1} 1 - \frac{1}{2}\ln 2) - 0 \). 5. Combine: \( I = 2\tan^{-1} 2 - \frac{1}{2}\ln 5 - \pi/4 + \frac{1}{2}\ln 2 - (\pi/4 - \frac{1}{2}\ln 2) \) \[ I = 2\tan^{-1} 2 - \frac{1}{2}\ln 5 + \ln 2 - \pi/2 = 2\tan^{-1} 2 - \frac{1}{2}\ln(5/4) - \pi/2 \].
Step 4: Final Answer:
The value is \( 2\tan^{-1} 2 - \frac{1}{2} \ln \frac{5}{4} - \frac{\pi}{2} \).