calculate-0-1-dt-1-t-2-3-

Question Number 40132 by maxmathsup by imad last updated on 16/Jul/18

c a l c u l a t e \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{3}}

Commented by maxmathsup by imad last updated on 21/Jul/18

l e t I = \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}}

I = \int_{0}^{1} \frac{1 + t^{2}}{{(1 + t^{2})}^{3}} d t = \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{3}} + \int_{0}^{1} \frac{t^{2}}{{(1 + t^{2})}^{3}} d t \Rightarrow

\int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{3}} = I - \int_{0}^{1} \frac{t^{2}}{{(1 + t^{2})}^{3}} c h a n g e m e n t t = t a n x g i v e

I = \int_{0}^{\frac{π}{4}} \frac{(1 + {t a n}^{2} x) d x}{{(1 + {t a n}^{2} x)}^{2}} = \int_{0}^{\frac{π}{4}} \frac{d x}{1 + {t a n}^{2} x} = \int_{0}^{\frac{π}{4}} {c o s}^{2} x d x

= \int_{0}^{\frac{π}{4}} \frac{1 + c o s (2 x)}{2} d x = \frac{π}{8} + \frac{1}{4} {[s i n (2 x)]}_{0}^{\frac{π}{4}} = \frac{π}{8} + \frac{1}{4} a l s o b y p a r t s

\int_{0}^{1} t {(1 + t^{2})}^{- 3} t d t = - \frac{1}{4} \int_{0}^{1} - 4 t {(1 + t^{2})}^{- 3} t d t

= - \frac{1}{4} {{[{(1 + t^{2})}^{- 2} t]}_{0}^{1} - \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}}}

= - \frac{1}{4} 2^{- 2} + \frac{1}{4} (\frac{π}{8} + \frac{1}{4}) = \frac{π}{32} + \frac{1}{16} - \frac{1}{16} = \frac{π}{32} \Rightarrow

\int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{3}} = \frac{π}{8} + \frac{1}{4} - \frac{π}{32} = \frac{3 π}{32} + \frac{1}{4} .