Question

4∫ dx/√12+4x-x² ,Evaluate it.0

Answer 1

we have to evaluate, $\int\limits^4_0{\frac{1}{\sqrt{12+4x-x^2}}}\,dx$

first of all resolve, 12 + 4x - x²

= -(x² - 4x - 12)

= -(x² - 4x + 4 - 16)

= -{(x - 2)² - (4)²}

= 4 - (x - 2)²

so, $\int\limits^4_0{\frac{1}{\sqrt{12+4x-x^2}}}\,dx=\int\limits^4_0{\frac{1}{\sqrt{4^2-(x-2)^2}}}\,dx$

we know,

$\int{\frac{1}{\sqrt{a^2-x^2}}}\,dx=sin^{-1}\left(\frac{x}{a}\right)+C$

so, $\int\limits^4_0{\frac{1}{\sqrt{4^2-(x-2)^2}}}\,dx=\left[sin^{-1}\left(\frac{(x-2)}{4}\right)\right]^4_0$

= sin-¹[(4-2)/4] - sin-¹[(0 - 2)/4]

= π/6 + π/6

= π/3