let-give-a-lt-1-find-the-value-of-f-a-0-pi-2-dx-1-acos-2-x-

Question Number 29446 by prof Abdo imad last updated on 08/Feb/18

l e t g i v e a < 1 f i n d t h e v a l u e o f

f (a) = \int_{0}^{\frac{π}{2}} \frac{d x}{1 - {a c o s}^{2} x} .

Commented by prof Abdo imad last updated on 07/Mar/18

f (a) = \int_{0}^{\frac{π}{2}} \frac{d x}{1 - a \frac{1 + c o s (2 x)}{2}} = \int_{0}^{\frac{π}{2}} \frac{2 d x}{2 - a - a c o s (2 x)}

= \int_{0}^{π} \frac{d t}{2 - a - a c o s t} a n d t h e c h . t a n (\frac{t}{2}) = u g i v e

f (a) = \int_{0}^{\infty} \frac{1}{2 - a - a \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{0}^{\infty} \frac{2 d u}{(2 - a) (1 + u^{2}) - a (1 - u^{2})}

= \int_{0}^{\infty} \frac{2 d u}{2 - a + (2 - a) u^{2} - a + {a u}^{2}}

= \int_{0}^{\infty} \frac{2 d u}{2 - 2 a + 2 u^{2}} = \int_{0}^{\infty} \frac{d u}{u^{2} - a + 1}

= \int_{0}^{\infty} \frac{d u}{u^{2} + 1 - a} t h e c h . u = \sqrt{1 - a} x g i v e

f (a) = \int_{0}^{\infty} \frac{\sqrt{1 - a} d x}{(1 - a) (1 + x^{2})} = \frac{1}{\sqrt{1 - a}} \frac{π}{2} \Rightarrow

f (a) = \frac{π}{2 \sqrt{1 - a}} .

Answered by sma3l2996 last updated on 09/Feb/18

t = t a n x \Rightarrow d x = \frac{d t}{1 + t^{2}}

f (a) = \int_{0}^{+ \infty} \frac{1}{1 - a \times \frac{1}{1 + t^{2}}} \times \frac{d t}{1 + t^{2}}

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

t = \sqrt{1 - a} u \Rightarrow d t = \sqrt{1 - a} d u

f (a) = \frac{\sqrt{1 - a}}{1 - a} \int_{0}^{\infty} \frac{d u}{u^{2} + 1} = \frac{\sqrt{1 - a}}{1 - a} {[a r c t a n u]}_{0}^{+ \infty}

f (a) = \frac{\sqrt{1 - a}}{1 - a} \times \frac{π}{2}