Q106 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
$\int \dfrac{1}{\sqrt{9+8 x-x^2}} d x=\varphi(x)+c$ then $\varphi(x)=$
- A$\dfrac{1}{10} \log \left|\dfrac{4-x}{4+x}\right|$
- Correct answerB$\sin ^{-1}\left(\dfrac{x-4}{5}\right)$
- C$\log \left|\dfrac{4-x}{4+x}\right|$
- D$\dfrac{1}{5} \sin ^{-1}\left(\dfrac{x-4}{5}\right)$
Solution
The integral is given by $I = \int \dfrac{1}{\sqrt{9+8x-x^2}} dx$.
Complete the square for the quadratic expression inside the square root:
$9 + 8x - x^2 = 9 - (x^2 - 8x) = 9 - (x^2 - 8x + 16 - 16) = 9 - ((x-4)^2 - 16) = 9 + 16 - (x-4)^2 = 25 - (x-4)^2$.
Substitute this into the integral:
$I = \int \dfrac{1}{\sqrt{25 - (x-4)^2}} dx$.
Using the standard integral formula $\int \dfrac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}\left(\dfrac{u}{a}\right) + C$, where $a = 5$ and $u = x-4$, we get:
$I = \sin^{-1}\left(\dfrac{x-4}{5}\right) + C$.
Thus, $\varphi(x) = \sin^{-1}\left(\dfrac{x-4}{5}\right)$.
Answer: $\sin ^{-1}\left(\dfrac{x-4}{5}\right)$
Complete the square for the quadratic expression inside the square root:
$9 + 8x - x^2 = 9 - (x^2 - 8x) = 9 - (x^2 - 8x + 16 - 16) = 9 - ((x-4)^2 - 16) = 9 + 16 - (x-4)^2 = 25 - (x-4)^2$.
Substitute this into the integral:
$I = \int \dfrac{1}{\sqrt{25 - (x-4)^2}} dx$.
Using the standard integral formula $\int \dfrac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}\left(\dfrac{u}{a}\right) + C$, where $a = 5$ and $u = x-4$, we get:
$I = \sin^{-1}\left(\dfrac{x-4}{5}\right) + C$.
Thus, $\varphi(x) = \sin^{-1}\left(\dfrac{x-4}{5}\right)$.
Answer: $\sin ^{-1}\left(\dfrac{x-4}{5}\right)$
Hence, the correct answer is option: 'B'.