let-give-x-lt-1-find-0-pi-2-d-1-x-2-cos-2-

Question Number 29972 by abdo imad last updated on 14/Feb/18

l e t g i v e ∣ x ∣< 1 f i n d \int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{1 - x^{2} {c o s}^{2} θ}} .

Commented by abdo imad last updated on 18/Feb/18

w e h a v e {(1 + u^{})}^{α} = 1 + α u + \frac{α (α - 1)}{2!} u^{2} + \dots

= 1 + \sum_{n = 1}^{\infty} \frac{α (α - 1) \dots (α - n + 1)}{n!} u^{n} w i t h ∣ u ∣< 1 w e h a v e

∣ x c o s θ ∣< 1 \Rightarrow \frac{1}{\sqrt{1 - x^{2} {c o s}^{2} θ}} = {(1 - {(x c o s θ)}^{2})}^{- \frac{1}{2}}

= 1 + \sum_{n = 1}^{\infty} \frac{(- \frac{1}{2}) (- \frac{1}{2} - 1) \dots (- \frac{1}{2} - n + 1)}{n!} x^{2 n} {c o s}^{2 n} θ l e t

p u t A_{n} = (- \frac{1}{2}) (- \frac{1}{2} - 1) \dots (- \frac{1}{2} - n + 1) \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{1 - x^{2} {c o s}^{2} θ}} = \frac{π}{2} + \sum_{n = 1}^{\infty} \frac{A_{n}}{n!} x^{2 n} \int_{0}^{\frac{π}{2}} {c o s}^{2 n} θ d θ b u t

w_{n} = \int_{0}^{\frac{π}{2}} {c o s}^{2 n} θ d θ i s k n o w n (w a l l i s i n t e g r a l) s o

\int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{1 - x^{2} {c o s}^{2} θ}} = \frac{π}{2} + \sum_{n = 1}^{\infty} \frac{A_{n} B_{n}}{n!} x^{2 n} .