0-a-a-2-x-2-a-2-x-2-2-dx-

Question Number 94312 by i jagooll last updated on 18/May/20

\underset{0}{\int^{a}} \frac{a^{2} - x^{2}}{{(a^{2} + x^{2})}^{2}} d x ?

Commented by mathmax by abdo last updated on 18/May/20

l e t I = \int_{0}^{a} \frac{a^{2} - x^{2}}{{(a^{2} + x^{2})}^{2}} d x \Rightarrow I =_{x = a t} \int_{0}^{1} \frac{a^{2} - a^{2} t^{2}}{a^{4} {(1 + t^{2})}^{2}} \times a d t

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

= \int_{0}^{1} \frac{d t}{1 + t^{2}} - 2 \int_{0}^{1} \frac{t^{2}}{{(1 + t^{2})}^{2}} d t w e h a v e \int_{0}^{1} \frac{d t}{1 + t^{2}} = {[a r c t a n t]}_{0}^{1} = \frac{π}{4}

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

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

= \int_{0}^{\frac{π}{4}} {c o s}^{2} u d u = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + c o s (2 u)) d u = \frac{π}{4} + \frac{1}{4} {[s i n (2 u)]}_{0}^{\frac{π}{4}}

= \frac{π}{4} + \frac{1}{4} \Rightarrow a I = \frac{π}{4} - 2 {\frac{π}{4} - \frac{π}{4} - \frac{1}{4}}

= \frac{π}{4} + \frac{1}{2} \Rightarrow I = \frac{1}{a} (\frac{π}{4} + \frac{1}{2}) w i t h a \neq 0

f o r a = 0 I i s n o t d e f i n e d!

Commented by mathmax by abdo last updated on 18/May/20

e r r o r o f t y p o \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}} = \frac{π}{8} + \frac{1}{4} \Rightarrow a I = \frac{π}{4} - 2 {\frac{π}{4} - \frac{π}{8} - \frac{1}{4}}

= \frac{π}{4} - \frac{π}{2} + \frac{π}{4} + \frac{1}{2} = \frac{1}{2} \Rightarrow I = \frac{1}{2 a}

Commented by i jagooll last updated on 18/May/20

yes . ===