PrepRung
Sign inSign up free

NDA 2025 Maths · Mathematics · Integrals · hard

Find the indefinite integral: ∫(sin x)/(1+cos ^2x) dx

Single correct option

Find the indefinite integral: ∫sin⁡x1+cos⁡2x dx\displaystyle\int\dfrac{\sin x}{1+\cos^2x}\,dx
  1. A. −tan⁡−1(cos⁡x)+C-\tan^{-1}(\cos x)+C
  2. B. tan⁡−1(cos⁡x)+C\tan^{-1}(\cos x)+C
  3. C. −ln⁡(1+cos⁡2x)+C-\ln(1+\cos^2x)+C
  4. D. tan⁡−1(sin⁡x)+C\tan^{-1}(\sin x)+C
Show answer and solution

Answer: (A) −tan⁡−1(cos⁡x)+C-\tan^{-1}(\cos x)+C

Put u=cos⁡xu=\cos x, du=−sin⁡x dxdu=-\sin x\,dx. The integral becomes −∫du1+u2=−tan⁡−1u+C=−tan⁡−1(cos⁡x)+C-\int\frac{du}{1+u^2}=-\tan^{-1}u+C=-\tan^{-1}(\cos x)+C. The option without the minus sign forgets du=−sin⁡x dxdu=-\sin x\,dx; the log form needs the numerator to be the derivative of the denominator, which here it is not.
Report a mistake

Original question written in the NDA 2025 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake