\(\int\cos^{-1}x\,dx\) is equal to (\(-1 \lt x \lt 1\), C is a constant of integration)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x\cos^{-1}x-\sqrt{1-x^2}+C\)
Writing the integrand as \(\cos^{-1}x \times 1\), we use integration by parts with \(\cos^{-1}x\) as the first function and 1 as the second.
\( \displaystyle \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}},\quad \int 1\,dx = x \)
So
\( \displaystyle \int\cos^{-1}x\,dx = x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^2}}\,dx \)
Putting \(t = 1-x^2\), \(dt = -2x\,dx\):
\( \displaystyle \int\frac{x}{\sqrt{1-x^2}}\,dx = -\frac{1}{2}\int t^{-1/2}\,dt = -\sqrt{1-x^2} \)
Hence
\( \int\cos^{-1}x\,dx = x\cos^{-1}x-\sqrt{1-x^2}+C \)
Why the other options are incorrect
Option (1) \(x\cos^{-1}x+\sqrt{1-x^2}+C\) — Incorrect. This sign belongs to \(\int\sin^{-1}x\,dx = x\sin^{-1}x+\sqrt{1-x^2}+C\); here the derivative of \(\cos^{-1}x\) is negative, which flips the sign.
Option (2) \(x\cos^{-1}x+x\sqrt{1-x^2}+C\) — Incorrect. At x = 0 its derivative is \(\cos^{-1}0+1 = \frac{\pi}{2}+1\), not \(\frac{\pi}{2}\).
Option (4) \(-x\cos^{-1}x+\sqrt{1-x^2}+C\) — Incorrect. Its derivative is \(-\cos^{-1}x\), the negative of the integrand.