Question-119006

Question Number 119006 by A8;15: last updated on 21/Oct/20

Answered by Dwaipayan Shikari last updated on 21/Oct/20

\int_{\frac{1}{2}}^{2} \frac{{t a n}^{- 1} x}{x^{2} - x + 1} d x = I

= \int_{2}^{\frac{1}{2}} \frac{{t a n}^{- 1} \frac{1}{t}}{\frac{1}{t^{2}} - \frac{1}{t} + 1} . \frac{d t}{- t^{2}} x = \frac{1}{t} \Rightarrow 1 = \frac{1}{- t^{2}} . \frac{d t}{d x}

= \int_{\frac{1}{2}}^{2} \frac{\frac{π}{2} - {t a n}^{- 1} t}{t^{2} - t + 1} d t = I

2 I = \int_{\frac{1}{2}}^{2} \frac{\frac{π}{2}}{t^{2} - t + 1} d t

4 I = π \int_{\frac{1}{2}}^{2} \frac{1}{t^{2} - t + 1} d t

\frac{4 I}{π} = \int_{\frac{1}{2}}^{2} \frac{1}{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d t

\frac{4 I}{π} = \frac{2}{\sqrt{3}} {[{t a n}^{- 1} \frac{2 t - 1}{\sqrt{3}}]}_{\frac{1}{2}}^{2}

I = \frac{π}{4} . \frac{2}{\sqrt{3}} . \frac{π}{3} = \frac{π^{2}}{6 \sqrt{3}}

Commented by A8;15: last updated on 21/Oct/20

thanks sir

Answered by Bird last updated on 22/Oct/20

w e p u t f o r a > 0 f (a) = \int_{\frac{1}{a}}^{a} \frac{a r c t a n x}{x^{2} - x + 1} d x

\Rightarrow f (a) =_{x = \frac{1}{t}} - \int_{\frac{1}{a}}^{a} \frac{\frac{π}{2} - a r c t a n t}{\frac{1}{t^{2}} - \frac{1}{t} + 1} (- \frac{d t}{t^{2}})

= \int_{\frac{1}{a}}^{a} \frac{\frac{π}{2} - a r c t a n t}{1 - t + t^{2}} d t

= \frac{π}{2} \int_{\frac{1}{a}}^{a} \frac{d t}{t^{2} - t + 1} - f (a) \Rightarrow

2 f (a) = \frac{π}{2} \int_{\frac{1}{a}}^{a} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}}

=_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} z} \frac{π}{2} \times \frac{4}{3} \int_{\frac{\frac{2}{a} - 1}{\sqrt{3}}}^{\frac{2 a - 1}{\sqrt{3}}} \frac{1}{1 + z^{2}} \times \frac{\sqrt{3}}{2} d z

= \frac{2 π}{3} \times \frac{\sqrt{3}}{2} {a r c t a n (\frac{2 a - 1}{\sqrt{3}}) - a r c t a n (\frac{\frac{2}{a} - 1}{\sqrt{3}})}

= \frac{π}{\sqrt{3}} {a r c t a n (\frac{2 a - 1}{\sqrt{3}}) - a r c t a n (\frac{\frac{2}{a} - 1}{\sqrt{3}})}

2 f (2) = \frac{π}{\sqrt{3}} {a r c t a n (\sqrt{3}) - a r c t a n (0)}

= \frac{π}{\sqrt{3}} \times \frac{π}{3} = \frac{π^{2}}{3 \sqrt{3}} \Rightarrow f (2) = \frac{π^{2}}{6 \sqrt{3}}

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