Math Question and Answers


Question and Answers Forum

All Questions Topic List
Vector Questions
Previous in All Question Next in All Question
Previous in Vector Next in Vector
Question Number 114808 by mnjuly1970 last updated on 21/Sep/20

$. . . n i c e m a t h . . .$ $e v a l u a t e ::$ $I = \int_{0}^{\frac{π}{4}} \frac{l n (1 + s i n (x))}{c o s (x)} d x = ? ? ?$ $. . . m . n . j u l y . 1970 . . .$
	Answered by mathdave last updated on 22/Sep/20

	$s o l u t i o n$ $l e t$ $I = \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \sin x)}{\cos x} d x = \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \sin x)}{1 - \sin^{2} x} \cos x d x$ $p u t y = \sin x$ $I = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\ln (1 + y)}{1 - y^{2}} d y = \frac{1}{2} (\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\ln (1 + y)}{1 - y} d y + \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\ln (1 + y)}{1 + y} d y) = \frac{1}{2} (A + B)$ $p u t y = 1 - 2 x i n A$ $l e t$ $A = \int_{\frac{1}{2} - \frac{\sqrt{2}}{4}}^{\frac{1}{\sqrt{2}}} \frac{\ln (2 - 2 x)}{x} d x = \int_{\frac{1}{2} - \frac{\sqrt{2}}{4}}^{\frac{1}{2}} (\frac{\ln (1 - x)}{x} + \frac{\ln 2}{x}) d x$ $A = \int_{\frac{1}{2} - \frac{\sqrt{2}}{4}}^{\frac{1}{2}} \frac{\ln (1 - x)}{x} d x + \ln 2 \int_{\frac{1}{2} - \frac{\sqrt{2}}{4}}^{\frac{1}{2}} \frac{1}{x} d x$ $A = {[- {L i}_{2} (x)]}_{\frac{1}{2} - \frac{\sqrt{2}}{4}}^{\frac{1}{2}} + \ln 2 {[\ln x]}_{\frac{1}{2} - \frac{\sqrt{2}}{4}}^{\frac{1}{2}}$ $A = - {L i}_{2} (\frac{1}{2}) + {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) + \ln 2 [\ln (\frac{1}{2}) - \ln (\frac{1}{2} - \frac{\sqrt{2}}{4})]$ $A = - \ln^{2} (2) - {L i}_{2} (\frac{1}{2}) + {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) - \ln 2 \ln (2 - \sqrt{2}) + {2 \ln}^{2} (2)$ $A = \frac{3}{2} \ln^{2} (2) - \frac{π^{2}}{12} + {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) - \ln 2 \ln (2 - \sqrt{2}) . . . . . (1) a n d$ $B = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\ln (1 + y)}{1 + y} d y u s i n g I B P$ $B = \ln^{2} (1 + y) ∣_{0}^{\frac{1}{\sqrt{2}}} - \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\ln (1 + y)}{1 + y} d y$ $B = \frac{1}{2} \ln^{2} (1 + \frac{1}{\sqrt{2}}) = \frac{1}{2} \ln^{2} (\frac{2 + \sqrt{2}}{2}) = \frac{1}{2} {[\ln (\frac{2 + \sqrt{2}}{2})]}^{2}$ $B = \frac{1}{2} {[\ln (2 + \sqrt{2}) - \ln 2]}^{2}$ $B = \frac{1}{2} \ln^{2} (2 + \sqrt{2}) + \frac{1}{2} \ln^{2} (2) - \ln 2 \ln (2 + \sqrt{2}) . . . . (2)$ $b u t I = \frac{1}{2} (A + B)$ $I = \frac{1}{2} (\frac{3}{2} \ln^{2} (2) - \frac{π^{2}}{12} + {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) - \ln 2 \ln (2 - \sqrt{2}) + \frac{1}{2} \ln^{2} (2 + \sqrt{2}) - \ln 2 \ln (2 + \sqrt{2}) + \frac{1}{2} \ln^{2} (2))$ $I = \frac{1}{2} ({2 \ln}^{2} (2) - \frac{π^{2}}{12} + {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) - \ln 2 [\ln (2 - \sqrt{2}) + \ln (2 + \sqrt{2})] + \frac{1}{2} \ln^{2} (2 + \sqrt{2}))$ $I = \frac{1}{2} ({2 \ln}^{2} (2) - \frac{π^{2}}{12} + {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) - \ln^{2} (2) + \frac{1}{2} \ln^{2} (2 + \sqrt{2}))$ $I = \frac{1}{2} {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) - \frac{π^{2}}{24} + \frac{1}{4} \ln^{2} (2 + \sqrt{2}) + \frac{1}{2} \ln^{2} (2)$ $∵ \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \sin x)}{\cos x} d x = \frac{1}{2} {L i}_{2} (\frac{1}{2} - \frac{\sqrt{2}}{4}) - \frac{π^{2}}{24} + \frac{1}{4} \ln^{2} (2 + \sqrt{2}) + \frac{1}{2} \ln^{2} (2)$ $b y m a t h d a v e (22 / 09 / 2020)$
	Commented by mnjuly1970 last updated on 22/Sep/20

	$t h a n k y o u s o m u c h m r d a v e . v e r y$ $n i c e . . .$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum