Calculus: integrate the function?

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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

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

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

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...!

?_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