Question

Evaluate the integral: \[ \int \operatorname{Cos}^{-1} \left( \frac{1-x^2}{1+x^2} \right) dx \]

• A. \( 2[x\operatorname{Tan}^{-1}x-\log\sqrt{1+x^2}] + C \)
• B. \( 2x\operatorname{Tan}^{-1}x+\log\sqrt{1+x^2} + C \)
• C. \( x\operatorname{Tan}^{-1}x+\log\sqrt{1-x^2} + C \)
• D. \( 2[\operatorname{Tan}^{-1}x-\log\sqrt{1+x^2}] + C \)

Hint:

For inverse trigonometric integrals, identify substitution patterns that simplify the expression before differentiation.

Answer 1

The Correct Option is A

Using trigonometric identities to rewrite the given function: \[ \operatorname{Cos}^{-1} \left( \frac{1-x^2}{1+x^2} \right) = 2\operatorname{Tan}^{-1} x \] Thus, the integral reduces to: \[ \int 2\operatorname{Tan}^{-1} x dx \] Applying integration by parts: \[ 2[x\operatorname{Tan}^{-1} x - \log \sqrt{1 + x^2}] + C \] Thus, the correct answer is: \[ 2[x\operatorname{Tan}^{-1}x-\log\sqrt{1+x^2}] + C \]