INT [(x+Sinx)/(1+cosx)]dx

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- Mar 27th 2007, 07:37 AMberkanatciI have a questions!!
INT [(x+Sinx)/(1+cosx)]dx

- Mar 27th 2007, 08:32 AMtopsquark
Divide this into two integrals:

Int[(x+Sinx)/(1+cosx) dx] = Int[x/(1 + cos(x)) dx] + Int[sin(x)/(1 + cos(x)) dx]

The first integral can be done by parts:

Int[p dq] = pq - Int[q dp]

Let p = x, dp = dx

Let dq = dx/(1 + cos(x)), q = Int[dx/(1 + cos(x))] = tan(x/2)

So

Int[x/(1 + cos(x)) dx] = x*tan(x/2) - Int[tan(x/2) dx] = x*tan(x/2) - (-2)*ln{|cos(x/2)|} = x*tan(x/2) + 2*ln{|cos(x/2)|}

The second integral may be done by substitution: y = cos(x), dy = -sin(x) dx:

Int[sin(x)/(1 + cos(x)) dx] = -Int[dy/(1 + y)] = -ln{|y + 1|} = -ln{|cos(x) + 1|}

Thus:

Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + 2*ln{|cos(x/2)|} - ln{|cos(x) + 1|}

Now, |cos(x/2)| = sqrt{(1 + cos(x))/2} so

ln{|cos(x/2)|} = ln[sqrt{(1 + cos(x))/2}] = ln[sqrt{(1 + cos(x))}] - ln[sqrt{2}] = (1/2)ln[cos(x) + 1] - (1/2)ln[2] so.....

Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + 2*ln{|cos(x/2)|} - ln{|cos(x) + 1|}

= x*tan(x/2) + ln{cos(x) + 1} - ln{2} - ln{|cos(x) + 1|}

= x*tan(x/2) - ln{2}

and, of course we need to add an arbitrary constant, so I'd say that

Int[(x+Sinx)/(1+cosx) dx] = x*tan(x/2) + C

-Dan - Mar 27th 2007, 08:50 AMecMathGeek
That x is not very easy to deal with. If you have a typo such that that x is actually supposed to be a 1, then this integration will be easy, but with an x I think it will take a few by-parts integrations to deal with.

**INT (x + sinx)/(1 + cosx) dx**

Let's multiply the numerator and denominator by 1 - cosx

INT [(x + sinx)(1 - cosx)]/(1 - cos^2x) dx

INT (x - x*cosx + sinx - sinxcosx)/sin^2x dx

Let's break this up into 4 fractions:

INT x/sin^2x dx - INT x*cosx/sin^2x dx + INT sinx/sin^2x - INT sinx*cosx/sin^2x dx

INT x*csc^2x dx - INT x*cotx*cscx dx + INT cscx dx - INT cotx dx

Now we will integrate each of these individually:

First:

INT x*csc^2x dx

Let u = x <--> du = dx

Let dv = csc^2x dx <--> v = -cotx

INT x*csc^2x dx = -xcotx + INT cotx dx = -xcotx + ln(abs(sinx))

Second:

INT x*cotx*cscx dx

Let u = x <--> du = dx

Let dv = cotx*cscx dx <--> v = -cscx

INT x*cotx*cscx dx = -x*cscx + INT cscx dx = -x*cscx - ln(abs(cscx + cotx))

Third:

INT cscx dx = -ln(abs(cscx + cotx))

Forth:

INT cotx dx = ln(abs(sinx))

Therefore:

INT (x + sinx)/(1 + cosx) dx = (-xcotx + ln(abs(sinx))) - (-x*cscx - ln(abs(cscx + cotx))) + (-ln(abs(cscx + cotx))) - (ln(abs(sinx))) +**C**

I don't feel like simplifying that :p ... Check to make sure I did the integrations correctly. I suspect I may have made a few mistakes.

- Mar 27th 2007, 08:55 AMecMathGeek
- Mar 27th 2007, 08:58 AMecMathGeek
- Mar 27th 2007, 08:59 AMberkanatci
- Mar 27th 2007, 10:32 AMtopsquark
(sigh) To be honest I did it on my calculator and it looked like a simple enough result I never went back to see if I could do it by hand. I CAN'T (baby-like wail). So I'll simplify your result down to mine:

csc(x) - cot(x) = 1/sin(x) - cos(x)/sin(x) = (1 - cos(x))/sin(x)

= (1 - cos(x))/sqrt(1 - cos^2(x)) = (1 - cos(x))/[sqrt(1 - cos(x))*sqrt(1 + cos(x))]

= sqrt(1 - cos(x))/sqrt(1 + cos(x)) = sqrt((1 - cos(x))/2)/sqrt((1 + cos(x))/2)

= sin(x/2)/cos(x/2) = tan(x/2)

-Dan