$\int_0^1 \frac{1}{\sqrt{3+2 x-x^2}} d x=$
$\int_0^1 \frac{1}{\sqrt{3+2 x-x^2}} d x=$
- $\frac{\pi}{2}$
- $\frac{\pi}{4}$
- $\frac{\pi}{3}$
- $\frac{\pi}{6}$
Solution
$\begin{aligned} & \int_0^1 \frac{1}{\sqrt{3+2 x-x^2}} \mathrm{~d} x=\int_0^1 \frac{\mathrm{d} x}{\sqrt{(2)^2-(x-1)^2}}=\left[\sin ^{-1}\left(\frac{x-1}{2}\right)\right]_0^1 \\ & =\sin ^{-1}(0)-\sin ^{-1}\left(\frac{-1}{2}\right)=0-\left(-\frac{\pi}{6}\right)=\frac{\pi}{6}\end{aligned}$
Asked in: MHT CET 2022 (10 Aug Shift 1)
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