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Question Number 27184 by abdo imad last updated on 02/Jan/18

c a l c u l a t e i n t e r m s o f x f (x) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x s i n t} .

Commented by abdo imad last updated on 05/Jan/18

l e t d o t h e c h a n g e m e n t t a n (\frac{t}{2}) = α \Leftrightarrow t = 2 a r c t a n α

f (x) = \int_{0}^{1} \frac{\frac{2 d α}{1 + α^{2}}}{1 + \frac{2 x α}{1 + α^{2}}} = \int_{0}^{1} \frac{2 d α}{1 + α^{2} + 2 x α} = \int_{0}^{1} \frac{2 d α}{α^{2} + 2 x α + 1}

= \int_{0}^{1} \frac{2 d α}{{(α + x)}^{2} + 1 - x^{2}}

c a s e 1 / x / < 1 w e d o t h e c h . α + x = \sqrt{1 - x^{2}} t

f (x) = \int_{\frac{x}{\sqrt{1 - x^{2}}}}^{\frac{x + 1}{\sqrt{1 - x^{2}}}} \frac{2 \sqrt{1 - x^{2}} d t}{(1 - x^{2}) (1 + t^{2})}

f (x) = \frac{2}{\sqrt{1 - x^{2}}} {[a r c t a n (t)]}_{\frac{x}{\sqrt{1 - x^{2}}}}^{\frac{\sqrt{1 + x}}{\sqrt{1 - x}}}

f (x) = \frac{2}{\sqrt{1 - x^{2}}} (a r c t a n (\frac{\sqrt{1 + x}}{\sqrt{1 - x}}) - a r c t a n (\frac{x}{\sqrt{1 - x^{2}}}))

c a s e 2 i f / x / > 1

f (x) = \frac{1}{x} \int_{0}^{\frac{π}{2}} \frac{d t}{x^{- 1} + s i n x} a n d w e c a n u s e t h e s a m e m t h o d e

d u e t o / x^{- 1} / < 1 \dots . .