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

Concept: Use the identity \[ \cot^{-1}x=\tan^{-1}\left(\frac{1}{x}\right) \] and then simplify using the property of inverse tangent.
Step 1: Convert to \(\tan^{-1}\) \[ \int_{0}^{1}\cot^{-1}(1+x+x^2)\,dx = \int_{0}^{1}\tan^{-1}\left(\frac{1}{1+x+x^2}\right)dx \] Rewrite \[ \frac{1}{1+x+x^2} = \frac{(x+1)-x}{1+x(x+1)} \] Using identity \[ \tan^{-1}A-\tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \] we obtain \[ \tan^{-1}\left(\frac{1}{1+x+x^2}\right) = \tan^{-1}(x+1)-\tan^{-1}x \]
Step 2: Split the integral \[ \int_{0}^{1}(\tan^{-1}(x+1)-\tan^{-1}x)\,dx \]
Step 3: Integrate Using \[ \int \tan^{-1}x\,dx=x\tan^{-1}x-\frac12\ln(1+x^2) \] we get \[ \left[x\tan^{-1}(x+1)-\frac12\ln(1+(x+1)^2)+\tan^{-1}(x+1)\right]_0^1 \] \[ -\left[x\tan^{-1}x-\frac12\ln(1+x^2)\right]_0^1 \]
Step 4: Substitute limits \[ =2\tan^{-1}2-\frac{\pi}{4}-\frac12\ln\left(\frac{5}{2}\right) -\left(\frac{\pi}{4}-\frac12\ln2\right) \] \[ =2\tan^{-1}2-\frac{\pi}{2}-\frac12\ln\left(\frac{5}{4}\right) \] Thus \[ \boxed{2\tan^{-1}2-\frac12\ln\left(\frac{5}{4}\right)-\frac{\pi}{2}} \]