f-x-0-pi-2-sin-2-t-1-xsin-2-t-dt-

Question Number 87052 by lémùst last updated on 02/Apr/20

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

Commented by mathmax by abdo last updated on 02/Apr/20

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

= \frac{π}{2 x} - \frac{1}{x} \int_{0}^{\frac{π}{2}} \frac{d t}{1 + {x s i n}^{2} t} [w e h a v e

\int_{0}^{\frac{π}{2}} \frac{d t}{1 + {x s i n}^{2} t} = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x \times \frac{1 - c o s (2 t)}{2}}

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

=_{t a n (\frac{u}{2}) = z} \int_{0}^{\infty} \frac{2 d z}{(1 + z^{2}) (2 + x - x \frac{1 - z^{2}}{1 + z^{2}})} = \int_{0}^{\infty} \frac{2 d z}{2 + x + (2 + x) z^{2} - x + {x z}^{2}}

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

c a s e 1 1 + x > 0 w e d o c h a n g e m e n t u = \sqrt{1 + x} z \Rightarrow

\int_{0}^{\infty} \frac{d z}{1 + (1 + x) z^{2}} = \int_{0}^{\infty} \frac{d u}{\sqrt{1 + x} (1 + u^{2})} = \frac{1}{\sqrt{1 + x}} \times \frac{π}{2} \Rightarrow

f (x) = \frac{π}{2 x} - \frac{π}{2 x \sqrt{1 + x}} (x > - 1 a n d x \neq 0)

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

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

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

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

= \frac{1}{2} {[\frac{1}{α} l n ∣ α z + 1 ∣ - \frac{1}{α} l n ∣ α z - 1 ∣]}_{0}^{+ \infty}

\frac{1}{2 α} {[l n ∣ \frac{α z + 1}{α z - 1} ∣]}_{0}^{+ \infty} = 0 s o

f (x) = \frac{π}{2 x} - \frac{π}{2 x \sqrt{1 + x}} i f x > - 1 a n d x \neq - 1

f (x) = 0 i f x < - 1

Commented by lémùst last updated on 02/Apr/20

m e r c i b e a u c o u p

m a i s j e p e n s e q u e f (x) = \frac{π}{2 x} s i x < - 1

Commented by mathmax by abdo last updated on 02/Apr/20

o u i t u a r a i s o n j a i o u b l i e \frac{π}{2 x} m e r c i .

Commented by Ar Brandon last updated on 03/Apr/20

g e n i a l!!