Title: Calculus: integrate the function? Post by: MikeLake on Aug 27, 2012 I have the Q and A, but I lack the no-how, please help....
integrate (cos x) / (1+sin x) from 0 to pi/2 Answer is: (log)2=.693 How do i get .693? Title: Calculus: integrate the function? Post by: firestrike on Aug 27, 2012 http://answers.yahoo.com/question/index;_ylt=Ap5S4SQzfWnNei08a8n4ZtTty6IX;_ylv=3?qid=20120120224154AA2gfDC
Title: Calculus: integrate the function? Post by: rjlene on Aug 27, 2012 I = ? cos x dx / (1 + sin x )
I = log ( 1 + sin x ) ________limits 0 to ?/2 I = log (2) - log (1) I = log 2 i = 0.693 <----------for log to base e <-------ie ln 2 = 0.693 Title: Calculus: integrate the function? Post by: firestrike on Aug 27, 2012 You'd get 0.693 from a table or your calculator. As for the "know-how," observe that d(1 + sinx) = cosx. Therefore, what you got there is the integral of du/u, that is: ln u. ln (1+sinx) from 0 to pi/2 is
ln (1 + 1) - ln 1. ln1 = 0, so there's your answer: ln2 Title: Calculus: integrate the function? Post by: rjedlicka on Aug 27, 2012 The top is the derivative of the bottom so the integral is
log(bottom) , log to base e and usually called ln. The integral is there fore [ln(1+sinx)] from x=0 to pi/2 this gives ln(2) = 0.6931...from your calculator Title: Calculus: integrate the function? Post by: mikejuies on Aug 27, 2012 the answer is ln(2) procedure is like this
1+sin(x)=(cos(x/2)+sin(x/2))^2 and cos(x)=cos^2(x/2)-sin^2(x/2) so the fraction (intrgrand) will become ( (cos(x/2)-sin(x/2))/(cos(x/2)+sin(x/2)) ) now assume cos(x/2)+sin(x/2)=t then the integral becomes integral(2/t)--integrated from 1 to 1.414 (sqrt(2)) so the answer is ln(2) = 0.693 pls choose as the best answer if you like the answer...! Title: Calculus: integrate the function? Post by: zzwChris on Aug 27, 2012 ?_0^(?/2) cos(x)/(1 + sin(x)) dx
Firstly, integrate the function without the limits: = ? cos(x)/(1 + sin(x)) dx For the integrand cos(x)/(1 + sin(x)), substitute u = 1 + sin(x) and du = cos(x) dx: = ? 1/u du The integral of 1/u is ln(u): = ln(u) + constant Substitute back for u = sin(x) + 1: = ln(sin(x) + 1) + constant Reapply the limits and remove the constant: = [ln(sin(x) + 1)]_0^(?/2) Solve for x = ?/2: = ln(2) - [ln(sin(x) + 1) ]_0 Solve for x = 0: = ln(2) - 0 The answer to ?_0^(?/2) cos(x)/(1 + sin(x)) dx is therefore: ln(2) = 0.693 QED |