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Question Number 32039 by abdo imad last updated on 18/Mar/18

a > - 1 c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d t}{1 + a {t a n}^{2} t} .

2) f i n d \int_{0}^{\frac{π}{2}} \frac{{t a n}^{2} t}{{(1 + {a t a n}^{2} t)}^{2}} d t

3) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{{t a n}^{2} t}{{(1 + 2 {t a n}^{2} t)}^{2}} d t .

Commented by abdo imad last updated on 22/Mar/18

l e t p u t f (a) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + {a t a n}^{2} t}

f (a) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + a \frac{{s i n}^{2} t}{{c o s}^{2} t}} = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} t}{{c o s}^{2} t + {a s i n}^{2} t} d t

f (a) = \int_{0}^{\frac{π}{2}} \frac{\frac{1 + c o s (2 t)}{2}}{\frac{1 + c o s (2 t}{2} + a \frac{1 - c o s (2 t)}{2}} d t

= \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 t)}{1 + c o s (2 t) + a (1 - c o s (2 t))} d t

= \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 t)}{1 + a + (1 - a) c o s (2 t)} t h e c h . 2 t = u g i v e

f (a) = \frac{1}{2} \int_{0}^{π} \frac{1 + c o s (u)}{1 + a + (1 - a) c o s (u)} d u c h . t a n (\frac{u}{2}) = x g i v e

f (a) = \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1 - x^{2}}{1 + x^{2}}}{1 + a + (1 - a) \frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}}

f (a) = \int_{0}^{\infty} \frac{2}{(1 + x^{2}) ((1 + a) (1 + x^{2}) + (1 - a) (1 - x^{2})} d x

= \int_{0}^{\infty} \frac{2 d x}{(1 + x^{2}) (1 + a + (1 + a) x^{2} + (1 - a) + (a - 1) x^{2})}

= \int_{0}^{\infty} \frac{2 d x}{(1 + x^{2}) (2 + 2 {a x}^{2})} = \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + {a x}^{2})}

= \frac{1}{1 - a} \int_{0}^{\infty} (\frac{1}{1 + x^{2}} - \frac{a}{1 + {a x}^{2}}) d x

f (a) = \frac{π}{2 (1 - a)} - \frac{a}{1 - a} \int_{0}^{\infty} \frac{d x}{1 + {a x}^{2}} b u t w e h a v e

b y c h . \sqrt{a} x = α (w e t a k e a > 0)

\int_{0}^{\infty} \frac{d x}{1 + {a x}^{2}} = \int_{0}^{\infty} \frac{1}{1 + α^{2}} \frac{d α}{\sqrt{a}} = \frac{π}{2 \sqrt{a}} \Rightarrow

f (a) = \frac{π}{2 (1 - a)} - \frac{a}{1 - a} \frac{π}{2 \sqrt{a}}

f (a) = \frac{π}{2 (1 - a)} (1 - \sqrt{a}) \Rightarrow f (a) = \frac{π}{2 (1 + \sqrt{a})} .

Commented by prof Abdo imad last updated on 22/Mar/18

2) w e h a v e p r o v e d t h a t

\int_{0}^{\frac{π}{2}} \frac{d t}{1 + {a t a n}^{2} t} = f (a) = \frac{π}{2 (1 + \sqrt{a})} l e t d e r i v a t e

f^{^{'}} (a) = - \int_{0}^{\frac{π}{2}} \frac{{t a n}^{2} t}{{(1 + a {t a n}^{2} t)}^{2}} d t \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{{t a n}^{2} t}{{(1 + a {t a n}^{2} t)}^{2}} d t = - f^{^{'}} (a) b u t

f^{^{'}} (a) = - \frac{π}{2} (\frac{{1 + \sqrt{a})}^{^{'}}}{{(1 + \sqrt{a})}^{2}}) = - \frac{π}{2} \frac{1}{2 \sqrt{a} {(1 + \sqrt{a})}^{2}} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{{t a n}^{2} t}{(1 + {a t a n}^{2} t)} d t = \frac{π}{4 \sqrt{a} {(1 + \sqrt{a})}^{2}}

3) a = 2 \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{{t a n}^{2} t}{(1 + 2 {t a n}^{2} t)} d t = \frac{π}{{4 \sqrt{2 (} 1 + \sqrt{2})}^{2}} .

Commented by abdo imad last updated on 22/Mar/18

a > 0 .