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

Question Number 66328 by mathmax by abdo last updated on 12/Aug/19

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

Commented by mathmax by abdo last updated on 15/Aug/19

l e t I = \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{3}} 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)}^{3}} d θ = \int_{0}^{\frac{π}{4}} \frac{d θ}{{(1 + {t a n}^{2} θ)}^{2}} = \int_{0}^{\frac{π}{4}} {c o s}^{4} θ d θ

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

= \frac{1}{8} \int_{0}^{\frac{π}{4}} {3 + 4 c o s (2 θ) + c o s (4 θ)} d θ

= \frac{3}{8} \frac{π}{4} + \frac{1}{2} \int_{0}^{\frac{π}{4}} c o s (2 θ) d θ + \frac{1}{8} \int_{0}^{\frac{π}{4}} c o s (4 θ) d θ

= \frac{3 π}{32} + \frac{1}{4} {[s i n (2 θ)]}_{0}^{\frac{π}{4}} + \frac{1}{32} {[s i n (4 θ)]}_{0}^{\frac{π}{4}}

= \frac{3 π}{32} + \frac{1}{4} + 0 \Rightarrow I = \frac{3 π}{32} + \frac{1}{4}