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Question Number 43823 by abdo.msup.com last updated on 15/Sep/18

l e t φ (a, x) = \int_{0}^{\frac{π}{2}} \frac{d t}{x + {a s i n}^{2} t}

2) f i n d a e x l i c i t e f o r m o f φ (a, x)

3) d e t e r m i n e φ (1, x) a n d φ (a, 1)

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

5) f i n d \int_{0}^{\frac{π}{2}} \frac{d t}{{(x + {a s i n}^{2} t)}^{2}}

6) f i n d \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} t}{{(x + a {s i n}^{2} t)}^{2}} d t .

Commented by maxmathsup by imad last updated on 17/Sep/18

1) w e h a v e φ (a, x) = \int_{0}^{\frac{π}{2}} \frac{d t}{x + a \frac{1 - c o s (2 t)}{2}} = \int_{0}^{\frac{π}{2}} \frac{2 d t}{2 x + a - a c o s (2 t)}

=_{2 t = u} \int_{0}^{π} \frac{d u}{2 x + a - a c o s u} =_{t a n (\frac{u}{2}) = α} \int_{0}^{\infty} \frac{1}{2 x + a - a \frac{1 - α^{2}}{1 + α^{2}}} \frac{2 d α}{1 + α^{2}}

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

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

c a s e 1 \frac{2 x + 1 - a}{2 x + 1 + a} > 0 c h a n g e m e n t α = \sqrt{\frac{2 x + 1 - a}{2 x + 1 + a}} z g i v e

φ (a, x) = \frac{2}{(2 x + 1 + a)} \int_{0}^{\infty} \frac{1}{\frac{2 x + 1 - a}{2 x + 1 + a} (1 + z^{2})} \sqrt{\frac{2 x + 1 - a}{2 x + 1 + a}} d z

= \frac{2}{2 x + 1 + a} \frac{2 x + 1 + a}{2 x + 1 - a} \sqrt{\frac{2 x + 1 - a}{2 x + 1 + a}} \frac{π}{2} = \frac{π}{2 x + 1 + a} \sqrt{\frac{2 x + 1 + a}{2 x + 1 - a}}

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

c a s e 2 \frac{2 x + 1 - a}{2 x + 1 + a} < 0 \Rightarrow φ (a, x) = \frac{2}{2 x + 1 + a} \int_{0}^{\infty} \frac{d α}{α^{2} - \frac{a - (2 x + 1)}{a + 2 x + 1}}

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

φ (a, x) = \frac{2}{2 x + 1 + a} \int_{0}^{\infty} \frac{1}{\frac{a - (2 x + 1)}{a + 2 x + 1} (z^{2} - 1)} \sqrt{\frac{a - (2 x + 1)}{a + (2 x + 1)}} d z

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

= \frac{π}{\sqrt{a^{2} - {(2 x + 1)}^{2}}} \frac{1}{2} \int_{0}^{\infty} {\frac{1}{z - 1} - \frac{1}{z + 1}} d z = \frac{π}{2 \sqrt{a^{2} - {(2 x + 1)}^{2}}} {[l n ∣ \frac{z - 1}{z + 1} ∣]}_{0}^{+ \infty} = 0

Commented by maxmathsup by imad last updated on 17/Sep/18

2) f o r a = 1 w e g e t \frac{2 x + 1 - a}{2 x + 1 + a} = \frac{2 x}{2 x + 2} = \frac{x}{x + 1}

c a s e 1 \frac{x}{x + 1} > 0 \Rightarrow φ (1, x) = \frac{π}{\sqrt{4 {(x + 1)}^{2} - 1}}

c a s e 2 \frac{x}{x + 1} < 0 \Rightarrow φ (1, x) = 0

a l s o f o r x = 1 w e h a v e \frac{2 x + 1 - a}{2 x + 1 + a} = \frac{3 - a}{3 + a}

c a s e 1 \frac{3 - a}{3 + a} > 0 \Rightarrow φ (a, 1) = \frac{π}{\sqrt{{(3 + a)}^{2} - a^{2}}}

c a s e 2 \frac{3 - a}{3 + a} < 0 \Rightarrow φ (a, 1) = 0 .

Commented by maxmathsup by imad last updated on 17/Sep/18

3) \int_{0}^{\frac{π}{2}} \frac{d t}{2 + 3 {s i n}^{2} t} = φ (3, 2) w e h a v e a = 3 a n d x = 2

w e g e t \frac{2 x + 1 - a}{2 x + 1 + a} = \frac{5 - 3}{8} = \frac{1}{4} > 0 \Rightarrow φ (3, 2) = \frac{π}{\sqrt{{(5 + 3)}^{2} - 3^{2}}} = \frac{π}{\sqrt{64 - 9}} = \frac{π}{\sqrt{55}}

Commented by maxmathsup by imad last updated on 18/Sep/18

4) w e h a v e φ (a, x) = \int_{0}^{\frac{π}{2}} \frac{d t}{x + a {s i n}^{2} t} \Rightarrow \frac{\partial φ}{\partial x} (a, x) = - \int_{0}^{\frac{π}{2}} \frac{d t}{{(x + {a s i n}^{2} t)}^{2}} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{d t}{{(x + {a s i n}^{2} t)}^{2}} = - \frac{\partial φ}{\partial x} (a, x)

c a s e 1 \frac{2 x + 1 - a}{2 x + 1 + a} > 0 \Rightarrow φ (a, x) = π {{(2 x + 1 + a)}^{2} - a^{2}}^{- \frac{1}{2}} \Rightarrow

\frac{\partial φ}{\partial x} (a, x) = - \frac{π}{2} (4 (2 x + 1 + a)) {{(2 x + 1 + a)}^{2} - a^{2}}^{- \frac{3}{2}}

= \frac{2 π (2 x + 1 + a)}{({(2 x + 1 + a)}^{2} - a^{2}) \sqrt{{(2 x + 1 + a)}^{2} - a^{2}}} .

c a s e 2 \frac{z x + 1 - a}{2 x + 1 + a} < 0 \Rightarrow φ (a, x) = 0 \Rightarrow \frac{\partial φ}{\partial x} (a, x) = 0 .