find-f-a-0-1-dt-a-2-t-2-3-with-a-gt-0-

Question Number 42793 by maxmathsup by imad last updated on 02/Sep/18

f i n d f (a) = \int_{0}^{1} \frac{d t}{{(a^{2} + t^{2})}^{3}} w i t h a > 0

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

t = a t a n α d t = {a s e c}^{2} α d α

\int_{0}^{{t a n}^{- 1} (\frac{1}{a})} \frac{{a s e c}^{2} α d α}{{a^{2} (1 + {t a n}^{2} α)}^{3}}

\int_{0}^{{t a n}^{- 1} (\frac{1}{a})} \frac{{a s e c}^{2} α d α}{a^{6} {s e c}^{6} α}

\frac{1}{a^{5}} \int_{0}^{{t a n}^{- 1} (\frac{1}{a})} {(\frac{1 + c o s 2 α}{2})}^{2} d α

\frac{1}{4 a^{5}} \int_{0}^{{t a n}^{- 1} (\frac{1}{a})} 1 + 2 c o s 2 α + \frac{1 + c o s 4 α}{2} d α

\frac{1}{8 a^{5}} \int_{0}^{{t a n}^{- 1} (\frac{1}{a})} 3 + 4 c o s 2 α + c o s 4 α d α

\frac{1}{8 a^{5}} ∣ (3 α + \frac{4 s i n 2 α}{2} + \frac{s i n 4 α}{4}) ∣_{0}^{{t a n}^{- 1} (\frac{1}{a})}

n o w t a n α = \frac{1}{a}

s i n 2 α = \frac{2 t a n α}{1 + {t a n}^{2} α} = \frac{\frac{2}{a}}{1 + \frac{1}{a^{2}}} = \frac{2 a}{1 + a^{2}}

s i n 4 α = 2 s i n 2 α . c o s 2 α

= 2 (\frac{2 t}{1 + t^{2}} \times \frac{1 - t^{2}}{1 + t^{2}})

= \frac{4 (\frac{1}{a}) (1 - \frac{1}{a^{2}})}{{(1 + \frac{1}{a^{2}})}^{2}} = \frac{\frac{4 (a^{2} - 1)}{a^{3}}}{\frac{{(a^{2} + 1)}^{2}}{a^{4}}} = \frac{4 a (a^{2} - 1)}{{(a^{2} + 1)}^{2}}

n o w p u t t h e v a l u e \dots

c o n t d s o t h e a n s i s

= \frac{1}{8 a^{5}} ∣ 3 {t a n}^{- 1} (\frac{1}{a}) + 2 . \frac{2 a}{1 + a^{2}} + \frac{1}{4} . \frac{4 a (a^{2} - 1)}{{(a^{2} + 1)}^{2}} ∣

= \frac{1}{8 a^{5}} {3 {t a n}^{- 1} (\frac{1}{a}) + \frac{4 a}{1 + a^{2}} + \frac{a^{3} - a}{{(a^{2} + 1)}^{2}}}