Math Question and Answers


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 109494 by mathdave last updated on 24/Aug/20

	Answered by mathmax by abdo last updated on 24/Aug/20

	$for ∣ u ∣< 1 we have \frac{d}{du} \ln (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} \Rightarrow$ $\ln (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} u^{n + 1} + c (c = 0) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} u^{n}}{n} \Rightarrow$ $\int_{0}^{\frac{π}{4}} \ln (1 + sinx) dx = \int_{0}^{\frac{π}{4}} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \sin^{n} xdx$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \int_{0}^{\frac{π}{4}} \sin^{n} x dx let A_{n} = \int_{0}^{\frac{π}{4}} \sin^{n} xdx$ $(wallis integral on [0, \frac{π}{4}]) \Rightarrow A_{n} = \int_{0}^{\frac{π}{4}} {(\frac{e^{ix} - e^{- ix}}{2 i})}^{n} dx$ $= \frac{1}{{(2 i)}^{n}} \int_{0}^{\frac{π}{4}} \sum_{k = 0}^{n} C_{n}^{k} {(e^{ix})}^{k} {(e^{- ix})}^{(n - k)}$ $= \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} C_{n}^{k} \int_{0}^{\frac{π}{4}} e^{ikx} e^{- i (n - k) x} dx$ $= \frac{1}{{(2 i)}^{n}} \sum_{k = 0}^{n} C_{n}^{k} \int_{0}^{\frac{π}{4}} e^{i (2 k - n) x} dx$ $= \frac{1}{(2 i)} \sum_{k = 0}^{n} C_{n}^{k} {[\frac{1}{i (2 k - n)} e^{i (2 k - n) x}]}_{0}^{\frac{π}{4}} = . . . . be continued . . .$
	Answered by mathmax by abdo last updated on 25/Aug/20
	$let try another way I =∫_0 ^(π/4) ln(cos^2 ((x/2))+sin^2 ((x/2))+2cos((x/2))sin((x/2))dx =∫_0 ^(π/4) ln{(cos((x/2))+sin((x/2))^2 )dx =2∫_0 ^(π/4) ln(cos((x/2))+sin((x/2)))dx =_((x/2)=t) 2 ∫_0 ^(π/8) ln(cost +sint)dt let f(a) =∫_0 ^(π/8) ln(cost +asint)dt f^′ (a) =∫_0 ^(π/8) ((sint)/(cost +a sint)) dt =∫_0 ^(π/8) (dt/(a+(1/(tant)))) =_(tant =u) ∫_0 ^((√2)−1) (du/((1+u^2 )(a+(1/u)))) =∫_0 ^((√2)−1) ((udu)/((au+1)(u^2 +1))) let decompose F(u) =(u/((au+1)(u^2 +1))) F(u) =(x/(au+1)) +((yu +z)/(u^2 +1)) ⇒x =((−1)/(a((1/a^2 )+1))) =((−1)/((1/a) +a)) =((−1)/(1+a^2 )) lim_(u→+∞) uF(u) =0 =(x/a) +y ⇒y =−(x/a) =(1/(a(1+a^2 ))) F(0) =0 =x +z ⇒z =−x =(1/(1+a^2 )) ⇒ F(u) =((−1)/((1+a^2 )(au+1))) +(((1/(a(1+a^2 )))u+(1/(1+a^2 )))/(u^2 +1)) =(1/(1+a^2 )){((−1)/(au+1)) +(1/a) (u/(u^2 +1)) +(1/(u^2 +1))} ⇒∫_0 ^((√2)−1) F(u)du =−(1/(1+a^2 ))∫_0 ^((√2)−1) (du/(au +1)) +(1/(2a(1+a^2 )))∫_0 ^((√2)−1) ((2udu)/(u^2 +1)) +(1/(1+a^2 ))∫_0 ^((√2)−1) (du/(1+a^2 )) =−(1/(a(1+a^2 )))[ln(au+1)]_0 ^((√2)−1) +(1/(2a(1+a^2 )))[ln(1+u^2 )]_0 ^((√2)−1) +(1/(1+a^2 ))arctan((√2)−1) =−((ln(a((√2)−1)+1))/(a(1+a^2 ))) +((ln(1+((√2)−1)^2 ))/(2a(1+a^2 ))) +(π/(8(1+a^2 ))) ⇒ ⇒f(a) =−∫ ((ln(a((√2)−1)+1))/(a(1+a^2 )))da +((ln(4−2(√2)))/2)arctana +(π/8) arctan(a) +C =−∫ ((ln(1+((√2)−1)a))/(a(1+a^2 )))da +((π/8) +((ln(4−2(√2)))/2))arctan(a) +C rest calculus of ∫ ((ln(1+((√2)−1)a))/(a(1+a^2 )))da....be continued...$
	$let try another way I = \int_{0}^{\frac{π}{4}} \ln (\cos^{2} (\frac{x}{2}) + \sin^{2} (\frac{x}{2}) + 2 \cos (\frac{x}{2}) \sin (\frac{x}{2}) dx$ $= \int_{0}^{\frac{π}{4}} \ln {(\cos (\frac{x}{2}) + \sin {(\frac{x}{2})}^{2}) dx = 2 \int_{0}^{\frac{π}{4}} \ln (\cos (\frac{x}{2}) + \sin (\frac{x}{2})) dx$ $=_{\frac{x}{2} = t} 2 \int_{0}^{\frac{π}{8}} \ln (cost + sint) dt let f (a) = \int_{0}^{\frac{π}{8}} \ln (cost + asint) dt$ $f^{^{'}} (a) = \int_{0}^{\frac{π}{8}} \frac{sint}{cost + a sint} dt = \int_{0}^{\frac{π}{8}} \frac{dt}{a + \frac{1}{tant}} =_{tant = u} \int_{0}^{\sqrt{2} - 1} \frac{du}{(1 + u^{2}) (a + \frac{1}{u})}$ $= \int_{0}^{\sqrt{2} - 1} \frac{udu}{(au + 1) (u^{2} + 1)} let decompose F (u) = \frac{u}{(au + 1) (u^{2} + 1)}$ $F (u) = \frac{x}{au + 1} + \frac{yu + z}{u^{2} + 1} \Rightarrow x = \frac{- 1}{a (\frac{1}{a^{2}} + 1)} = \frac{- 1}{\frac{1}{a} + a} = \frac{- 1}{1 + a^{2}}$ $\lim_{u \to + \infty} uF (u) = 0 = \frac{x}{a} + y \Rightarrow y = - \frac{x}{a} = \frac{1}{a (1 + a^{2})}$ $F (0) = 0 = x + z \Rightarrow z = - x = \frac{1}{1 + a^{2}} \Rightarrow$ $F (u) = \frac{- 1}{(1 + a^{2}) (au + 1)} + \frac{\frac{1}{a (1 + a^{2})} u + \frac{1}{1 + a^{2}}}{u^{2} + 1}$ $= \frac{1}{1 + a^{2}} {\frac{- 1}{au + 1} + \frac{1}{a} \frac{u}{u^{2} + 1} + \frac{1}{u^{2} + 1}} \Rightarrow \int_{0}^{\sqrt{2} - 1} F (u) du$ $= - \frac{1}{1 + a^{2}} \int_{0}^{\sqrt{2} - 1} \frac{du}{au + 1} + \frac{1}{2 a (1 + a^{2})} \int_{0}^{\sqrt{2} - 1} \frac{2 udu}{u^{2} + 1} + \frac{1}{1 + a^{2}} \int_{0}^{\sqrt{2} - 1} \frac{du}{1 + a^{2}}$ $= - \frac{1}{a (1 + a^{2})} {[\ln (au + 1)]}_{0}^{\sqrt{2} - 1} + \frac{1}{2 a (1 + a^{2})} {[\ln (1 + u^{2})]}_{0}^{\sqrt{2} - 1} + \frac{1}{1 + a^{2}} \arctan (\sqrt{2} - 1)$ $= - \frac{\ln (a (\sqrt{2} - 1) + 1)}{a (1 + a^{2})} + \frac{\ln (1 + {(\sqrt{2} - 1)}^{2})}{2 a (1 + a^{2})} + \frac{π}{8 (1 + a^{2})} \Rightarrow$ $\Rightarrow f (a) = - \int \frac{\ln (a (\sqrt{2} - 1) + 1)}{a (1 + a^{2})} da + \frac{\ln (4 - 2 \sqrt{2})}{2} arctana + \frac{π}{8} \arctan (a) + C$ $= - \int \frac{\ln (1 + (\sqrt{2} - 1) a)}{a (1 + a^{2})} da + (\frac{π}{8} + \frac{\ln (4 - 2 \sqrt{2})}{2}) \arctan (a) + C$ $rest calculus of \int \frac{\ln (1 + (\sqrt{2} - 1) a)}{a (1 + a^{2})} da . . . . be continued . . .$
	Commented by mathmax by abdo last updated on 25/Aug/20

	$sorry f (a) = - \int \frac{\ln (1 + (\sqrt{2} - 1) a)}{a (1 + a^{2})} da + \frac{\ln (4 - 2 \sqrt{2})}{2} \int \frac{da}{a (1 + a^{2})}$ $+ \frac{π}{8} \arctan (a) + C . . . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum