Mathematics

Evaluate
$$\displaystyle \int^{2}_{0}\dfrac {1}{\sqrt {3+2x-x^{2}}}dx$$


SOLUTION
Given,

$$\displaystyle \int _0^2 \dfrac{1}{\sqrt{3+2x-x^2}}dx$$

$$=\displaystyle\int _0^2\dfrac{1}{\sqrt{-\left(x-1\right)^2+4}}dx$$

substitute $$u=x-1$$ and   $$du=dx$$

$$=\displaystyle\int _{-1}^1\dfrac{1}{\sqrt{-u^2+4}}du$$

substitute $$u=2\sin v$$

$$=\displaystyle\int _{-\tfrac{\pi }{6}}^{\tfrac{\pi }{6}}1dv$$

$$=\left[v\right]^{\tfrac{\pi }{6}}_{-\tfrac{\pi }{6}}$$

$$=\dfrac{\pi }{3}$$
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Subjective Medium Published on 17th 09, 2020
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