Question

Find : \( \int \tan^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \, dx \)

Hint:

The substitution \(x = \cos \theta\) simplifies many algebraic fractional roots inside inverse trig functions. Always verify the range of the parameter when performing substitutions involving square roots and trig functions.

Answer 1

Concept:
• Trigonometric substitution is effective for expressions like \(\sqrt{\frac{1-x}{1+x}}\). Letting \(x = \cos \theta\) is ideal.
• Use half-angle formulas: \(1 - \cos \theta = 2\sin^2(\theta/2)\) and \(1 + \cos \theta = 2\cos^2(\theta/2)\).
• Integration by parts: \( \int u \, dv = uv - \int v \, du \).

Step 1: Substitute to simplify the inverse trigonometric function
Let \(x = \cos \theta\), then \(dx = -\sin \theta \, d\theta\). Also, \(\theta = \cos^{-1} x\). \[ \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} = \sqrt{\tan^2(\theta/2)} = \tan(\theta/2) \] The integral becomes: \[ I = \int \tan^{-1}(\tan(\theta/2)) \cdot (-\sin \theta) \, d\theta \] \[ I = -\frac{1}{2} \int \theta \sin \theta \, d\theta \]

Step 2: Integrate by parts
Let \(u = \theta\) and \(dv = \sin \theta \, d\theta\). Then \(du = d\theta\) and \(v = -\cos \theta\). \[ I = -\frac{1}{2} \left[ \theta(-\cos \theta) - \int (-\cos \theta) \, d\theta \right] \] \[ I = -\frac{1}{2} [-\theta \cos \theta + \sin \theta] + C \] \[ I = \frac{1}{2} \theta \cos \theta - \frac{1}{2} \sin \theta + C \]

Step 3: Back-substitute in terms of \(x\)
We have \(x = \cos \theta\), \(\theta = \cos^{-1} x\), and \(\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}\). \[ I = \frac{1}{2} x \cos^{-1} x - \frac{1}{2} \sqrt{1 - x^2} + C \]