Evaluate-0-1-2-dx-1-x-2-1-x-2-

Question Number 16703 by Tinkutara last updated on 25/Jun/17

Evaluate : \int_{0}^{\frac{1}{2}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}}

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17

x = t g φ \Rightarrow d x = (1 + {t g}^{2} φ) d φ

1 + x^{2} = \frac{1}{{c o s}^{2} φ} \Rightarrow {c o s}^{2} φ = \frac{1}{1 + x^{2}} \Rightarrow

{s i n}^{2} φ = 1 - \frac{1}{1 + x^{2}} = \frac{x^{2}}{1 + x^{2}} \Rightarrow s i n φ = \frac{x}{\sqrt{1 + x^{2}}}

\Rightarrow I = \int \frac{(1 + {t g}^{2} φ) d φ}{(1 + {t g}^{2} φ) \sqrt{1 - {t g}^{2} φ}} = \int \frac{d φ}{\sqrt{1 - {t g}^{2} φ}} =

= \int \frac{c o s φ d φ}{\sqrt{1 - 2 {s i n}^{2} φ}} = \int \frac{d u}{\sqrt{1 - 2 u^{2}}} = \frac{1}{\sqrt{2}} \int \frac{d u}{\sqrt{\frac{1}{2} - u^{2}}} =

= \frac{1}{\sqrt{2}} \times {s i n}^{- 1} \sqrt{2} u + C = \frac{1}{\sqrt{2}} {s i n}^{- 1} (\sqrt{2} s i n φ) + C =

= \frac{1}{\sqrt{2}} {s i n}^{- 1} (\frac{x \sqrt{2}}{\sqrt{1 + x^{2}}}) + C

\Rightarrow I = F (\frac{1}{2}) - F (0) = \frac{1}{\sqrt{2}} {s i n}^{- 1} (\frac{2 \times \frac{1}{2} \sqrt{2}}{\sqrt{5}}) - 0 =

= \frac{1}{\sqrt{2}} {s i n}^{- 1} (\sqrt{\frac{2}{5}}) = 0.483924 .

Commented by Tinkutara last updated on 26/Jun/17

Thanks Sir!

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17

y o u a r e r i g h t . i t i s c o r r e c t n o w .

Answered by Tinkutara last updated on 05/Jul/17

Let x = \frac{1}{t} \Rightarrow t = \frac{1}{x} and d x = \frac{- d t}{t^{2}}

∴ \int \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}} = \int \frac{\frac{- d t}{t^{2}}}{(1 + \frac{1}{t^{2}}) \sqrt{1 - \frac{1}{t^{2}}}}

= - \int \frac{t d t}{(t^{2} + 1) \sqrt{t^{2} - 1}}

Now let t^{2} - 1 = z^{2} . So t d t = z d z .

Integral becomes - \int \frac{z d z}{(z^{2} + 2) z}

= - \int \frac{d z}{z^{2} + 2} = \frac{- 1}{\sqrt{2}} \tan^{- 1} \frac{z}{\sqrt{2}}

Now z = \sqrt{t^{2} - 1} = \sqrt{\frac{1}{x^{2}} - 1}

As x \to 0, z \to \infty and x \to \frac{1}{2}, z \to \sqrt{3} .

∴ \int_{0}^{\frac{1}{2}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}} = \frac{- 1}{\sqrt{2}} {[\tan^{- 1} \frac{z}{\sqrt{2}}]}_{\infty}^{\sqrt{3}}

= \frac{1}{\sqrt{2}} [\frac{π}{2} - \tan^{- 1} \sqrt{\frac{3}{2}}]

= \frac{1}{\sqrt{2}} \tan^{- 1} \sqrt{\frac{2}{3}}