Question

\( \int x \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \, dx \, (x>0) = \)

• A. \(-x + (1+x^2) \tan^{-1} x + c\)
• B. \(x-(1+x^2) \cot^{-1} x + c\)
• C. \(-x + (1+x^2) \cot^{-1} x + c\)
• D. \(-x - (1+x^2) \tan^{-1} x + c\)

Hint:

When faced with integrals involving inverse trigonometric functions composed with rational expressions, use trigonometric identities like \( \frac{1-x^2}{1+x^2} = \cos 2 \theta \) to simplify before integration.

Answer 1

The Correct Option is A

We are asked to evaluate: \[ \int x \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \, dx \] Step 1: Recognize inverse trigonometric identity We know: \[ \frac{1-x^2}{1+x^2} = \cos 2\theta \] So, \[ 2\theta = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] Then, \[ \theta = \frac{1}{2} \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] And also, we know: \[ \tan \theta = x \] Step 2: Use integration by parts Let \[ I = \int x \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \, dx \] Let \[ u = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), \, dv = x\,dx \] Then, \[ du = \frac{4x}{(1+x^2)^2 \sqrt{1 - \left(\frac{1-x^2}{1+x^2}\right)^2}} \, dx \] But easier via substitution: \[ \text{Since } \frac{1-x^2}{1+x^2} = \cos 2 \theta, \text{ so } 2 \theta = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] \[ \text{And } \tan \theta = x \] \[ \Rightarrow \theta = \tan^{-1} x \] Therefore, \[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2 \tan^{-1} x \] Step 3: Substitute back Now, \[ I = \int x \times 2 \tan^{-1} x \, dx \] \[ = 2 \int x \tan^{-1} x \, dx \] Step 4: Apply integration by parts again Let \[ u = \tan^{-1} x, \, dv = x\,dx \] Then, \[ du = \frac{1}{1+x^2} \, dx, \, v = \frac{x^2}{2} \] Now, \[ I = 2 \left( \frac{x^2}{2} \tan^{-1} x - \int \frac{x^2}{2} \times \frac{1}{1+x^2} \, dx \right) \] \[ = 2 \left( \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx \right) \] Step 5: Simplify integral \[ \int \frac{x^2}{1+x^2} \, dx = \int \left(1 - \frac{1}{1+x^2}\right) \, dx \] \[ = x - \tan^{-1} x \] Step 6: Final substitution Now, \[ I = 2 \left( \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} (x - \tan^{-1} x) \right) \] \[ = x^2 \tan^{-1} x - (x - \tan^{-1} x) \] \[ = (x^2 + 1) \tan^{-1} x - x \] Finally, \[ \boxed{I = -x + (1 + x^2) \tan^{-1} x + c} \]