Homework Help

What is definite integral? integral sign low limit 0, upper limit pie/2...

user profile pic

lemong | Student, Undergraduate | (Level 1) Honors

Posted March 30, 2013 at 7:50 AM via web

dislike 1 like

What is definite integral?

integral sign low limit 0, upper limit pie/2 ln(1+sinx)/ln(1+sinxcosx+sinx+cosx)dx?

2 Answers | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 30, 2013 at 8:37 AM (Answer #1)

dislike 1 like

You need to convert ` 1 + sin xcos x + sin x + cos x` into its factored form, such that:

`1 + sin xcos x + sin x + cos x = (1 + sin x) + cos x*(1 + sin x)`

`1 + sin xcos x + sin x + cos x = (1 + sin x)(1 + cos x)`

Taking the logarithm yields:

`ln(1 + sin xcos x + sin x + cos x) = ln ((1 + sin x)(1 + cos x))`

Using logarithmic identites yields:

`ln(1 + sin xcos x + sin x + cos x) = ln(1 + sin x) + ln(1 + cos x)`

Replacing `ln(1 + sin x) + ln(1 + cos x) for ln(1 + sin xcos x + sin x + cos x)` under integral yields:

`int_0^(pi/2) (ln(1 + sin x))/(ln(1 + sin x) + ln(1 + cos x)) dx`

You need to add and subtract `ln(1 + cos x)` both sides, yields:

`int_0^(pi/2) (ln(1 + sin x) + ln(1 + cos x) - ln(1 + cos x))/(ln(1 + sin x) + ln(1 + cos x)) dx`

Splitting the integral yields:

`int_0^(pi/2) dx - int_0^(pi/2) (ln(1 + cos x))/(ln(1 + sin x) + ln(1 + cos x)) dx`

You need to use the following substitution, such that:

`x = 0 + pi/2 - t => 1 + cos x = 1 + cos(pi/2 - t) = 1 + sin t`

`1 + sin x = 1 + cos t`

`dx = -dt`

Performing the indicated substitutions, yields:

`int_0^(pi/2) (ln(1 + sin x))/(ln(1 + sin x) + ln(1 + cos x)) dx = -int_(pi/2)^0 (ln(1 + cos t))/(ln(1 + cos t) + ln(1 + sin t)) dt`

Using the fundamental theorem of calculus, yields:

`int_0^(pi/2) (ln(1 + sin x))/(ln(1 + sin x) + ln(1 + cos x)) dx = (pi/2 - 0)/2 = pi/4`

Hence, evaluating the given definite integral, using the indicated substitution, yields `int_0^(pi/2) (ln(1 + sin x))/(ln(1 + sin x) + ln(1 + cos x)) dx = (pi/2 - 0)/2 = pi/4.`

user profile pic

rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted March 30, 2013 at 10:24 AM (Answer #2)

dislike 1 like

Let I=`int_0^(pi/2)ln(1+sinx)/ln(1+sinxcosx+sinx+cosx)dx`

     = `int_0^(pi/2)ln(1+sinx)/ln{(1+sinx)(1+cosx)}dx`

  =   `int_0^(pi/2)ln(1+sinx)/{ln(1+sinx)+ln(1+cosx)}dx`      (I)

Now by rule `int_0^af(x)dx=int_0^af(a-x)dx`

So,  I=`int_0^(pi/2)ln(1+sin((pi/2)-x))/{ln(1+sin((pi/2)-x))+ln(1+cos((pi/2)-x))}dx`

or,  I=`int_0^(pi/2)ln(1+cosx)/{ln(1+cosx)+ln(1+sinx)}dx`      (II)

Now adding (I) and (II) we get

 2I=`int_0^(pi/2){ln(1+sinx)+ln(1+cosx)}/{ln(1+sinx)+ln(1+cosx)}dx`

or   2I=`int_0^(pi/2)1dx`

or,   2I=[x]   with lower limit 0 and upper limit `(pi/2)`

or,     2I=`(pi/2)-0`

or,    2I=`(pi/2)`

or,      I= `(pi/4)`         Answer.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes