\(\int\sqrt{8+2x-x^2}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{x-1}{2}\sqrt{8+2x-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x-1}{3}\right)+C\)
Completing the square:
\( 8+2x-x^2 = 8-(x^2-2x+1)+1 \)
\( = 3^2-(x-1)^2 \)
With \(t = x-1\), this is the standard integral
\( \displaystyle \int\sqrt{a^2-t^2}\,dt = \frac{t}{2}\sqrt{a^2-t^2}+\frac{a^2}{2}\sin^{-1}\frac{t}{a}+C \)
With a = 3, \(\frac{a^2}{2} = \frac{9}{2}\):
\( \displaystyle \int\sqrt{8+2x-x^2}\,dx = \frac{x-1}{2}\sqrt{8+2x-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x-1}{3}\right)+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{x-1}{2}\sqrt{8+2x-x^2}-\frac{9}{2}\sin^{-1}\left(\frac{x-1}{3}\right)+C\) — Incorrect. The two terms of the standard result are added; with a minus sign the derivative is not the integrand.
Option (2) \(\displaystyle\frac{x+1}{2}\sqrt{8+2x-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x+1}{3}\right)+C\) — Incorrect. The square is \((x-1)^2\), since \(x^2-2x = (x-1)^2-1\).
Option (4) \(\displaystyle\frac{x-1}{2}\sqrt{8+2x-x^2}+9\sin^{-1}\left(\frac{x-1}{3}\right)+C\) — Incorrect. The inverse sine term carries \(\frac{a^2}{2} = \frac{9}{2}\), not \(a^2 = 9\).