Question

integration of:
(x+sin x) / ( 1+cos x)
= (x + 2 sin x/2 cos x/2 ) / ( 2 co^2 x/2)
what is the final answer for this?

Answer 1

your answer is here: http://www.wolframalpha.com/input/?i=+integration+of%3A+%28x%2Bsin+x%29+%2F+%28+1%2Bcos+x%29+
however, if you want to do it..
http://en.wikipedia.org/wiki/Weierstrass_substitution
this might work

Answer 2

that will work

Answer 3

any better methods?
im too tired to think :/

Answer 4

if i proceed this way, i dont get the answer as: x tan x/2 + C
instead i get (x tan x/2) /2 - log(sec x/2)/2 + C, I am not sure where i went wrong can pls correct me

Answer 5

what method did you use?
\[\int \frac{x+sinx}{1+cosx}dx => \int \frac{x}{1+cosx}dx+\int\frac{sinx}{1+cosx}dx\]
might be easier..weistrauss sub then u-sub

Answer 6

I wonder how well parts would work. \[
u= x+\sin(x), dv = \frac{dx}{1+\cos(x)}
\]

Answer 7

When I look at that integral, it makes me think of quotient rule a bit for some reason.

Answer 8

the denominator is the derivative of the numerator

Answer 9

\[\int\limits_{}^{} \frac{ x }{ 2 \cos^2 x/2 } dx + \int\limits_{}^{} \frac{ 2 \sin x/2 \cos x/2 }{ 2 \cos^2 x/2 } dx\]

Answer 10

\[=\frac{ 1 }{ 2 } \int\limits_{}^{} x \sec^2 x/2 dx + \int\limits_{}^{} \tan x/2 dx \]

Answer 11

use the product rule for the first and keep the second as it is
\[\frac{ 1 }{ 2 } \left[ x \int\limits_{}^{} \sec^2 x/2 - \int\limits_{}^{} d/dx (x) \int\limits_{}^{} \sec^2 x/2 dx dx \right] + \int\limits_{}^{} \tan x/2 dx\]

Answer 12

\[= \frac{ 1 }{ 2 } \left[ \frac{ x \tan x/2 }{ 1/2 } - \int\limits_{}^{} \frac{ \tan x/2 }{ 1/2 } dx \right] +\int\limits_{}^{} \tan x/2 dx\]

Answer 13

so last two terms get cancelled and we left out with only
x tan x/2 + C

Answer 14

question, why did you write integral d/dx (x)

Answer 15

oh nevermind, maybe you should have used parenthesees, to be more clear. but very nice. how in the world did you think of this