prove-that-0-pi-xsin-x-1-cos-2-x-pi-2-4-

Question Number 76716 by peter frank last updated on 29/Dec/19

p r o v e t h a t

\int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} = \frac{π^{2}}{4}

Commented by abdomathmax last updated on 30/Dec/19

l e t I = \int_{0}^{π} \frac{x s i n x}{1 + {c o s}^{2} x} d x c h a n g e m e n t x = π - t g i v e

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

2 I = π \int_{0}^{π} \frac{s i n t}{1 + {c o s}^{2} t} d t w e h a v e

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

= 4 \int_{0}^{\infty} \frac{u}{{(1 + u^{2})}^{2} (1 + \frac{{(1 - u^{2})}^{2}}{{(1 + u^{2})}^{2}})} d u

= 4 \int_{0}^{\infty} \frac{u d u}{{(1 + u^{2})}^{2} + {(1 - u^{2})}^{2}}

= 4 \int_{0}^{\infty} \frac{u d u}{u^{4} + 2 u^{2} + 1 + u^{4} - 2 u^{2} + 1}

= 4 \int_{0}^{\infty} \frac{u d u}{2 u^{4} + 2} = 2 \int_{0}^{\infty} \frac{u d u}{u^{4} + 1}

=_{u = z^{\frac{1}{4}}} 2 \int_{0}^{\infty} \frac{z^{\frac{1}{4}}}{1 + z} \times \frac{1}{4} z^{\frac{1}{4} - 1} d z

= \frac{1}{2} \int_{0}^{\infty} \frac{z^{\frac{1}{2} - 1}}{1 + z} d z = \frac{1}{2} \times \frac{π}{s i n (\frac{π}{2})} = \frac{π}{2} \Rightarrow

2 I = \frac{π^{2}}{2} \Rightarrow ★ I = \frac{π^{2}}{4} ★

Commented by peter frank last updated on 30/Dec/19

t h a n k y o u

Commented by Tony Lin last updated on 30/Dec/19

\int_{0}^{π} \frac{x s i n x}{1 + {c o s}^{2} x} d x

l e t π - θ = x, d θ = - d x

\int_{0}^{π} x f (s i n x) d x

= \int_{π}^{0} (π - θ) f [s i n (π - θ)] (- d θ)

= \int_{0}^{π} (π - θ) f [s i n θ] d θ

= π \int_{0}^{π} f (s i n θ) d θ - \int_{0}^{π} θ f (s i n θ) d θ

= π \int_{0}^{π} f (s i n x) d x - \int_{0}^{π} x f (s i n x) d x

\Rightarrow \int_{0}^{π} x f (s i n x) d x = \frac{π}{2} \int_{0}^{π} f (s i n x) d x

\int_{0}^{π} \frac{x s i n x}{1 + {c o s}^{2} x} d x

= \frac{π}{2} \int_{0}^{π} \frac{s i n x}{1 + {c o s}^{2} x} d x

l e t c o s x = t, d t = - s i n x d x

\frac{π}{2} \int_{- 1}^{1} \frac{1}{1 + t^{2}} d t

= π {t a n}^{- 1} t ∣_{0}^{1}

= \frac{π^{2}}{4}

Commented by peter frank last updated on 30/Dec/19

t h a n k y o u

Answered by Kunal12588 last updated on 30/Dec/19

I = \int_{0}^{π} \frac{x s i n x}{1 + {c o s}^{2} x} d x

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

[u s i n g \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x]

a d d i n g r e d o n e s

2 I = π \int_{0}^{π} \frac{s i n x}{1 + {c o s}^{2} x} d x

(\begin{matrix} t = c o s x \Rightarrow d t = - s i n x d x \\ x \to 0 \Rightarrow t \to 1 & x \to π \Rightarrow t \to - 1 \end{matrix})

2 I = - π \int_{1}^{- 1} \frac{d t}{1 + t^{2}} = π \int_{- 1}^{1} \frac{d t}{1 + t^{2}} = 2 π \int_{0}^{1} \frac{d t}{1 + t^{2}}

[u s i n g - \int_{b}^{a} f (x) d x = \int_{a}^{b} f (x) d x]

[a n d \underset{- a}{\int^{a}} f (x) d x = {\begin{cases} \int_{0}^{2 a} f (x) d x; i f f (x) i s e v e n f u n c t i o n \\ 0; i f f (x) i s o d d f u n c t i o n \end{cases}]

\Rightarrow I = π {[{t a n}^{- 1} t]}_{0}^{1} = π [\frac{π}{4} - 0] = \frac{π^{2}}{4}

∴ \int_{0}^{π} \frac{x s i n x}{1 + {c o s}^{2} x} d x = \frac{π^{2}}{4} Q E D

Commented by peter frank last updated on 30/Dec/19

t h a n k y o u