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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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