calculus-Please-Evaluate-0-pi-2-x-sin-x-2-dx-M-N-july-1970-

Question Number 112579 by mnjuly1970 last updated on 08/Sep/20

\dots . c a l c u l u s \dots .

P l e a s e E v a l u a t e :::

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

You can't use 'macro parameter character #' in math mode

Answered by mathmax by abdo last updated on 09/Sep/20

Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{\sin^{2} x} dx \Rightarrow Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{1 - \cos^{2} x} dx we do the changement

tanx = t \Rightarrow x = \arctan (t) \Rightarrow Ω = \int_{0}^{\infty} \frac{\arctan^{2} t}{1 - \frac{1}{1 + t^{2}}} \times \frac{dt}{1 + t^{2}}

= \int_{0}^{\infty} \frac{\arctan^{2} t}{1 + t^{2} - 1} dt = \int_{0}^{\infty} \frac{\arctan^{2} t}{t^{2}} dt by parts

\int_{0}^{\infty} \frac{\arctan^{2} t}{t^{2}} dt = {[- \frac{\arctan^{2} t}{t}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{t} \times \frac{2 arctant}{(1 + t^{2})} dt

= 2 \int_{0}^{\infty} \frac{\arctan (t)}{t (1 + t^{2})} dt = \int_{- \infty}^{+ \infty} \frac{arctant}{t (t^{2} + 1)} dt let φ (z) = \frac{arctanz}{z (z^{2} + 1)}

\Rightarrow φ (z) = \frac{\arctan (z)}{z (z - i) (z + i)} residus theorem give

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, o) + Res (φ, i)}

Res (φ, o) = \lim_{z \to 0} z φ (z) = \lim_{z \to 0} \frac{arctanz}{z^{2} + 1} = 0

Res (φ, i) = \lim_{z \to i} (z - i) φ (z) = \frac{\arctan (i)}{i (2 i)} = - \frac{1}{2} \arctan (i)

\Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times (- \frac{1}{2} \arctan (i)}

= - i π \arctan (i) \Rightarrow Ω = - i π \arctan (i) rest to find \arctan (i)

\dots be continued

we have the formula arctanz = \frac{1}{2 i} \ln (\frac{1 + iz}{1 - iz}) but this is not

applicable for i \dots . . be continued \dots .

Commented by mnjuly1970 last updated on 09/Sep/20

g r a t e f u l s i r m a t h m a x . g o d

k e e p y o u . .

Commented by mathmax by abdo last updated on 10/Sep/20

thank you sir

Answered by mathdave last updated on 09/Sep/20

m y s o l u t i o n t o

I = \int_{0}^{\frac{π}{2}} {(\frac{x}{\sin x})}^{2} d x (u s i n g I B P)

I = {[- \frac{x^{2} \cos x}{\sin x}]}_{0}^{\frac{π}{2}} + 2 \int_{0}^{\frac{π}{2}} x \cot x d x (u s i n g I B P)

I = 0 + 2 {[x \ln (\sin x)]}_{0}^{\frac{. π}{2}} - 2 \int_{0}^{\frac{π}{2}} \ln (\sin x) d x

I = 0 + 0 - 2 \int_{0}^{\frac{π}{2}} \ln (\sin x) d x

b u t \int_{0}^{\frac{π}{2}} \ln (\sin x) d x = - \frac{π}{2} \ln 2

∵ I = - 2 [- \frac{π}{2} \ln 2] = π \ln 2

∵ \int_{0}^{\frac{π}{2}} {(\frac{x}{\sin x})}^{2} d x = π \ln 2

b y m a t h d a v e (08 / 09 / 2020)

Commented by mnjuly1970 last updated on 09/Sep/20

b r a v o b r a v o b r a v o s i r

p e a c e b e u p o n y o u \dots m r

d a v e

Commented by Tawa11 last updated on 06/Sep/21

great sir

Answered by maths mind last updated on 09/Sep/20

l e t t g (\frac{x}{2}) = t \Rightarrow d x = \frac{2 d t}{1 + t^{2}}

Ω = \int_{0}^{1} \frac{4 {a r c t a n}^{2} (t)}{{(\frac{2 t}{1 + t^{2}})}^{2}} . \frac{2 d t}{(1 + t^{2})} = 2 \int_{0}^{1} (1 + t^{2}) \frac{{a r c t a n}^{2} (t) d t}{t^{2}}

= 2 \int_{0}^{1} \frac{{a r c t a n}^{2} (t)}{t^{2}} {d t}_{= 2 I} + 2 \int_{0}^{1} {a r c t a n}^{2} (t) {d t}_{= 2 J}

B y P a r t

J = {[{t a r c t a n}^{2} (t)]}_{0}^{1} - \int \frac{2 t}{1 + t^{2}} a r c t a n (t) d t = \frac{π^{2}}{16} - \int \frac{2 t a r c t a n (t) d t}{1 + t^{2}}

= \frac{π^{2}}{16} - {[l n (1 + t^{2}) a r c t a n (t)]}_{0}^{1} + \int_{0}^{1} \frac{l n (1 + t^{2})}{1 + t^{2}} d t

= \frac{π^{2}}{16} - l n (2) \frac{π}{4} + \int_{0}^{1} \frac{l n (1 + t^{2})}{1 + t^{2}} d t

t = t g (δ) \Rightarrow \int_{0}^{1} \frac{l n (1 + t^{2})}{1 + t^{2}} d t = \int_{0}^{\frac{π}{4}} l n (1 + {t g}^{2} (δ)) d δ

w e u s e \int_{0}^{\frac{π}{2}} l n (c o s (x)) d x = - \frac{π l n (2)}{2} \Rightarrow

- 2 \int_{0}^{\frac{π}{2}} l n (c o s (δ)) d δ = - 2 . \frac{- π}{2} l n (2) = π l n (2)

2 \int_{0}^{\frac{π}{4}} l n (t g (t)) d t = - 2 G = - 2 \int_{0}^{\frac{π}{4}} l n (c o s (δ)) d δ + 2 \int_{\frac{π}{4}}^{\frac{π}{2}} l n (c o s (δ)) d δ

= - 2 \int_{0}^{\frac{π}{4}} l n (c o s (δ)) d δ - π l n (2) - 2 \int_{0}^{\frac{π}{4}} l n (c o s (δ)) d δ \Rightarrow - 2 \int_{0}^{\frac{π}{4}} l n (c o s (δ) d δ

= - G + \frac{π l n (2)}{2}

\Rightarrow J = \frac{π^{2}}{16} + \frac{π l n (2)}{4} - G

I b y p a r t = \lim_{x \to 0} {[- \frac{1}{t} {a r c t a n}^{2} (t)]}_{x}^{1} + \int_{0}^{1} \frac{2 a r c t a n (t)}{t (1 + t^{2})} d t

- \frac{π^{2}}{16} + 2 \int_{0}^{1} (\frac{1}{t} - \frac{t}{1 + t^{2}}) a r c t a n (t) d t

= - \frac{π^{2}}{16} + 2 \int_{0}^{1} \frac{a r c t a n (t)}{t} d \underset{= 2 G}{t} - \int_{0}^{1} \frac{2 t}{1 + t^{2}} a r c r a n (t) {d o n e l i g n (5))

= - \frac{π^{2}}{16} + 2 G + l n (2) \frac{π}{4} - G = - \frac{π^{2}}{16} + G + \frac{π l n (2)}{4}

2 I + 2 J = π l n (2)

G c a t a l a n c a n s t a n t

E \int_{0}^{1} \frac{a r c t a n (t)}{t} d t = \int_{0}^{1} \frac{1}{t} \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k} t^{2 k + 1}}{2 k + 1} d t = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}} = G . . t r u e

b y p a r t E = {[l n (t) a r c t a n (t)]}_{0}^{1} - \int_{0}^{1} \frac{l n (t)}{1 + t^{2}} d t

= - \int_{0}^{1} \frac{l n (t)}{1 + t^{2}} d t (t = t g (δ)) \Rightarrow= - \int_{0}^{\frac{π}{4}} l n (t g (δ)) d δ p r o v e d w h a t

w e u s e d i n l i g n / 11

Commented by mnjuly1970 last updated on 09/Sep/20

t h a n k y o u s o m u c h .

Answered by mnjuly1970 last updated on 09/Sep/20

m y s o l u t i o n .

\int_{0}^{\frac{π}{2}} l o g (s i n (x)) d x = - \frac{π}{2} l o g (2)

l . h . s \overset{i . b . p}{=} {[x l o g (s i n (x))]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} x c o t (x) d x

\overset{{l i m}_{x \to 0^{+}} (x l o g (s i n (x))) = 0}{=} - \int_{0}^{\frac{π}{2}} x c o t (x) d x

\overset{i . b . p . A g a i n}{=} - {{[\frac{x^{2}}{2} c o t (x)]}_{0}^{\frac{π}{2}} + \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{s i n}^{2} (x)} d x =

= - \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{s i n}^{2} (x)} d x = r . h . s = - \frac{π}{2} l o g (2)

∴ \int_{0}^{\frac{π}{2}} {(\frac{x}{s i n (x)})}^{2} d x = π l o g (2) ▴

You can't use 'macro parameter character #' in math mode