∫_( 0) ^(π/2) ln (2 cos x) dx = ?


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 77464 by naka3546 last updated on 06/Jan/20

$\int_{0} \overset{\frac{π}{2}}{} \ln (2 \cos x) d x = ?$
Commented by kaivan.ahmadi last updated on 06/Jan/20

$u = l n (2 c o s x) \Rightarrow d u = - t g x d x$ $d v = d x \Rightarrow v = x$ $u v - \int v d u = x l n (2 c o s x) + \int x t g x d x$ $n o w f o r \int x t g x d x$ $u = t g x \Rightarrow d u = (1 + {t g}^{2} x) d x$ $d v = x d x \Rightarrow v = \frac{x^{2}}{2}$ $u v - \int v d u = \frac{x^{2}}{2} t g x - \frac{1}{2} \int x^{2} (1 + {t g}^{2} x) d x =$ $\frac{x^{2}}{2} t g x - \frac{1}{2} \int (x^{2} + x^{2} {t g}^{2} x) d x$ $s o$
Commented by MJS last updated on 06/Jan/20

$sorry but this is wrong$ $I {don}^{'} t think {it}^{'} s possible to solve this using$ $elementary calculus$
Commented by kaivan.ahmadi last updated on 06/Jan/20

$i f F (x) = x l n (2 c o s x) + \frac{1}{2} (x^{2} t g x - x + {t g}^{- 1} x)$ $t h e n$ $F^{'} (x) = l n (2 c o s x)$ $s o \int l n (2 c o s x) d x = F (x) + C$
Commented by MJS last updated on 06/Jan/20

$but$ $\frac{d}{d x} [x \ln (2 \cos x) + \frac{1}{2} (x^{2} \tan x - x + \tan x)] =$ $= \ln (2 \cos x) + \frac{1}{2} (\tan^{2} x + \frac{x^{2}}{\cos^{2} x})$
Commented by mathmax by abdo last updated on 06/Jan/20

$\int_{0}^{\frac{π}{2}} l n (2 c o s x) d x = \frac{π}{2} l n (2) + \int_{0}^{\frac{π}{2}} l n (c o s x) d x w e h a v e p r o v e d t h a t$ $\int_{0}^{\frac{π}{2}} l n (c o s x) d x = - \frac{π}{2} l n (2) \Rightarrow \int_{0}^{\frac{π}{2}} l n (2 c o s x) d x = 0$
Commented by mathmax by abdo last updated on 06/Jan/20

$l e t f (a) = \int_{0}^{\frac{π}{2}} l n (a c o s x) d x w i t h a > 0 w e h a v e$ $f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{c o s x}{a c o s x} d x = \frac{π}{2 a} \Rightarrow f (a) = \frac{π}{2} l n (a) + c$ $f (1) = \int_{0}^{\frac{π}{2}} l n (c o s x) d x = - \frac{π}{2} l n (2) = c \Rightarrow$ $\int_{0}^{\frac{π}{2}} l n (a c o s x) d x = \frac{π}{2} l n (a) - \frac{π}{2} l n (2) \Rightarrow$ $\int_{0}^{\frac{π}{2}} l n (2 c o s x) d x = 0$
Commented by kaivan.ahmadi last updated on 06/Jan/20

$y e s s i r y o u a r e r i g h t . i f i n d m i s t a k e . i t h a s$ $\int x^{2} {t g}^{2} x d x$ $c a n w e s o l v e i t ?$
	Answered by MJS last updated on 06/Jan/20

	$\int \ln (2 \cos x) d x =$ $[t = - i x \to d x = i d t]$ $= - i \int t d t + i \int \ln (e^{2 t} + 1) d t$ $\int \ln (e^{2 t} + 1) d t =$ $[u = - e^{2 t} \to d t = - \frac{d x}{{2 e}^{2 x}}]$ $= \frac{1}{2} \int \frac{\ln (1 - u)}{u} d u = - \frac{1}{2} \int - \frac{\ln (1 - u)}{u} d u =$ $= - \frac{1}{2} {Li}_{2} u$ ${Li}_{2} x is the dilogarithm with {Li}_{2} x = \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n^{2}}$
	Answered by Kunal12588 last updated on 07/Jan/20
	$I=∫_0 ^( π/2) log(2 cos x) dx .....(1) ⇒I=∫_0 ^( π/2) log(2 cos ((π/2)−x))dx ⇒I=∫_0 ^( π/2) log(2 sin x) dx .....(2) adding (1) and (2) 2I=∫_0 ^( π/2) log(2 sin 2x) dx let 2x = t ⇒ { ((x→0 ⇒ t→0)),((x→ (π/2) ⇒ t→π)) :} ⇒dx=(1/2)dt ∴ 2I=(1/2)∫_0 ^( π) log(2 sin t) dt = (1/2)∫_0 ^( π) log(2 sin x)dx ⇒2I=(1/2)×2∫_0 ^( π/2) log(2 sin x)dx ⇒2I=∫_0 ^( π/2) log(2 sin ((π/2)−x))dx ⇒2I=∫_0 ^( π/2) log(2 cos x)dx ⇒2I=I ⇒I=0 ∴ ∫_0 ^( π/2) log(2 cos x)dx = 0$
	$I = \int_{0}^{π / 2} l o g (2 c o s x) d x . . . . . (1)$ $\Rightarrow I = \int_{0}^{π / 2} l o g (2 c o s (\frac{π}{2} - x)) d x$ $\Rightarrow I = \int_{0}^{π / 2} l o g (2 s i n x) d x . . . . . (2)$ $a d d i n g (1) a n d (2)$ $2 I = \int_{0}^{π / 2} l o g (2 s i n 2 x) d x$ $let 2 x = t \Rightarrow {\begin{cases} x \to 0 \Rightarrow t \to 0 \\ x \to \frac{π}{2} \Rightarrow t \to π \end{cases}$ $\Rightarrow d x = \frac{1}{2} d t$ $∴ 2 I = \frac{1}{2} \int_{0}^{π} l o g (2 s i n t) d t = \frac{1}{2} \int_{0}^{π} l o g (2 s i n x) d x$ $\Rightarrow 2 I = \frac{1}{2} \times 2 \int_{0}^{π / 2} l o g (2 s i n x) d x$ $\Rightarrow 2 I = \int_{0}^{π / 2} l o g (2 s i n (\frac{π}{2} - x)) d x$ $\Rightarrow 2 I = \int_{0}^{π / 2} l o g (2 c o s x) d x$ $\Rightarrow 2 I = I$ $\Rightarrow I = 0$ $∴ \int_{0}^{π / 2} l o g (2 c o s x) d x = 0$
	Answered by Kunal12588 last updated on 07/Jan/20

	$I = \int_{0}^{π / 2} l o g (a c o s x) d x$ $\Rightarrow I = \int_{0}^{π / 2} l o g (a s i n x) d x$ $\Rightarrow 2 I = \int_{0}^{π / 2} l o g (a^{2} s i n x c o s x) d x$ $\Rightarrow 2 I = \int_{0}^{π / 2} l o g (\frac{a^{2}}{2} s i n 2 x) d x$ $\Rightarrow 2 I = l o g (\frac{a^{2}}{2}) \int_{0}^{π / 2} d x + \int_{0}^{π / 2} l o g (s i n 2 x) d x$ $\Rightarrow 2 I = \frac{π}{2} l o g (\frac{a^{2}}{2}) + \frac{1}{2} \int_{0}^{π} l o g (s i n x) d x$ $\Rightarrow 2 I = \frac{π}{2} l o g (\frac{a^{2}}{2}) + \frac{1}{2} \times 2 \int_{0}^{π / 2} l o g (s i n x) d x$ $\Rightarrow 2 I = \frac{π}{2} l o g (\frac{a^{2}}{2}) + \int_{0}^{π / 2} l o g (c o s x) d x$ $\Rightarrow 2 I = \frac{π}{2} l o g (\frac{a^{2}}{2}) - \frac{π}{2} l o g (2)$ $\Rightarrow I = \frac{π}{4} l o g (\frac{a^{2}}{2}) - \frac{π}{4} l o g (2)$ $\Rightarrow I = \frac{π}{4} l o g (\frac{a^{2}}{4}) = \frac{π}{2} l o g (\frac{a}{2})$ $f (a) = \int_{0}^{π / 2} l o g (a c o s x) d x = \frac{π}{2} l o g (\frac{a}{2})$ $f (1) = \frac{π}{2} l o g (\frac{1}{2}) = - \frac{π}{2} l o g (2)$ $f (2) = \frac{π}{2} l o g (1) = 0$ $f (3) = \frac{π}{2} l o g (\frac{3}{2})$ $f (4) = \frac{π}{2} l o g (2)$ $⋮$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum