find-0-1-dt-1-t-2-2-

Question Number 40133 by maxmathsup by imad last updated on 16/Jul/18

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

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

c h a n g e m e n t t = t a n θ g i v e

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

= \int_{0}^{\frac{π}{4}} {c o s}^{2} θ d θ = \frac{1}{2} \int_{0}^{\frac{π}{4}} (1 + c o s (2 θ) d θ

= \frac{π}{8} + \frac{1}{2} {[\frac{1}{2} s i n (2 θ)]}_{0}^{\frac{π}{4}} = \frac{π}{8} + \frac{1}{4} \Rightarrow

I = \frac{π}{8} + \frac{1}{4}