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Question Number 41848 by maxmathsup by imad last updated on 13/Aug/18

l e t f (a) = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + a s i n x} w i t h a \in R

1) f i n d a s i m p l e f o r m o f f (a)

2) c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d x}{1 + s i n x} a n d \int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 s i n x}

3) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{c o s x}{{(1 + a s i n x)}^{2}} d x

4) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{c o s x}{{(1 + s i n x)}^{2}} d x a n d \int_{0}^{\frac{π}{2}} \frac{c o s x}{{(1 + 2 s i n x)}^{2}} d x

Commented by maxmathsup by imad last updated on 14/Aug/18

1) c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

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

r o o t s o f t^{2} + 2 a t + 1

Δ^{^{'}} = a^{2} - 1

c a s e 1 a^{2} > 1 \Rightarrow ∣ a ∣> 1 \Rightarrow t_{1} = - a + \sqrt{a^{2} - 1} q n d t_{2} = - a - \sqrt{a^{2} - 1} a n d

F (t) = \frac{2}{t^{2} + 2 a t + 1} = \frac{2}{(t - t_{1}) (t - t_{2})} = \frac{α}{t - t_{1}} + \frac{β}{t - t_{2}}

α = \frac{2}{t_{1} - t_{2}} = \frac{2}{2 \sqrt{a^{2} - 1}} = \frac{1}{\sqrt{a^{2} - 1}} a n d β = \frac{2}{t_{2} - t_{1}} = - \frac{1}{\sqrt{a^{2} - 1}} \Rightarrow

F (t) = \frac{1}{\sqrt{a^{2} - 1}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} \Rightarrow f (a) = \int_{0}^{1} \frac{1}{\sqrt{a^{2} - 1}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} d t

= \frac{1}{\sqrt{a^{2} - 1}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{1} = \frac{1}{\sqrt{a^{2} - 1}} {l n ∣ \frac{1 + a - \sqrt{a^{2} - 1}}{1 + a + \sqrt{a^{2} - 1}} ∣ - l n ∣ \frac{a - \sqrt{a^{2} - 1}}{a + \sqrt{a^{2} - 1}} ∣}

c a s e 2 a^{2} < 1 \Rightarrow - 1 < a < 1 \Rightarrow n o r e a l r o o t s \Rightarrow

f (a) = \int_{0}^{1} \frac{2 d t}{t^{2} + 2 a t + a^{2} + 1 - a^{2}} = \int_{0}^{1} \frac{2 d t}{{(t + a)}^{2} + 1 - a^{2}} c h a n g e m e n t

t + a = \sqrt{1 - a^{2}} u g i v e f (a) = \int_{\frac{a}{\sqrt{1 - a^{2}}}}^{\frac{1 + a}{\sqrt{1 - a^{2}}}} \frac{2 \sqrt{1 - a^{2}} d u}{(1 - a^{2}) (1 + u^{2})}

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

Commented by maxmathsup by imad last updated on 14/Aug/18

2) \int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 s i n x} = f (2) = \frac{1}{\sqrt{3}} {l n ∣ \frac{3 - \sqrt{3}}{3 + \sqrt{3}} ∣ - l n ∣ \frac{2 - \sqrt{3}}{2 + \sqrt{3}} ∣}

= \frac{1}{\sqrt{3}} {l n (\frac{3 - \sqrt{3}}{3 + \sqrt{3}}) - l n (\frac{2 - \sqrt{3}}{2 + \sqrt{3}})} l e t c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d x}{1 + s i n x}

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

= \int_{0}^{1} \frac{2 d u}{{(u + 1)}^{2}} = {[\frac{- 2}{1 + u}]}_{0}^{1} = - 2 (\frac{1}{2} - 1) = 1 .

Commented by maxmathsup by imad last updated on 15/Aug/18

3) w e h a v e f (a) = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + a s i n x} \Rightarrow f^{^{'}} (a) = - \int_{0}^{\frac{π}{2}} \frac{s i n x}{{(1 + a s i n x)}^{2}} d x

\Rightarrow \int_{0}^{\frac{π}{2}} \frac{s i n x}{{(1 + a s i n x)}^{2}} d x = - f^{^{'}} (a)

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

f^{^{'}} (a) = 2 (- \frac{- a}{\sqrt{1 - a^{2}}}) \times \frac{1}{(1 - a^{2})} {a r c t a n (\frac{1 + a}{\sqrt{1 - a^{2}}}) - a r c t a n (\frac{a}{\sqrt{1 - a^{2}}})}

+ \frac{2}{\sqrt{1 - a^{2}}} {\frac{{(\frac{1 + a}{\sqrt{1 - a^{2}}})}^{^{'}}}{1 + {(\frac{1 + a}{\sqrt{1 - a^{2}}})}^{2}} - \frac{{(\frac{a}{\sqrt{1 - a^{2}}})}^{^{'}}}{1 + {(\frac{a}{\sqrt{1 - a^{2}}})}^{2}}} b u t

{(\frac{1 + a}{\sqrt{1 - a^{2}}})}^{^{'}} = \frac{\sqrt{1 - a^{2}} - (1 + a) (\frac{- 2 a}{2 \sqrt{1 - a^{2}}})}{1 - a^{2}} = \frac{1 - a^{2} + a + a^{2}}{(1 - a^{2}) \sqrt{1 - a^{2}}} = \frac{1 + a}{(1 - a^{2}) \sqrt{1 - a^{2}}}

{(\frac{a}{\sqrt{1 - a^{2}}})}^{^{'}} = \frac{\sqrt{1 - a^{2}} - a (\frac{- a}{\sqrt{1 - a^{2}}})}{(1 - a^{2})} = \frac{1 - a^{2} + a^{2}}{(1 - a^{2}) \sqrt{1 - a^{2}}} = \frac{1}{(1 - a^{2}) \sqrt{1 - a^{2}}} \Rightarrow

f^{^{'}} (a) = \frac{2 a}{(1 - a^{2}) \sqrt{1 - a^{2}}} {a r c t a n (\frac{1 + a}{\sqrt{1 - a^{2}}}) - a r c t a n (\frac{a}{\sqrt{1 - a^{2}}})}

+ \frac{2}{\sqrt{1 - a^{2}}} {\frac{1 + a}{(1 - a^{2}) \sqrt{1 - a^{2}} (1 + \frac{{(1 + a)}^{2}}{1 - a^{2}})} - \frac{1}{(1 - a^{2}) \sqrt{1 - a^{2}} (1 + \frac{a^{2}}{1 - a^{2}})}}

i f ∣ a ∣> 1 w e f o l l o w t h e s a m e w a y .