Ytukyg

I have to evaluate the following integral by parts: $$int {dfrac{x+sin{x}}{1+cos{x}}}mathrm{d}x $$

So I tried to put:

$ u = x + sin{x}$ $~qquadrightarrow quad$ $mathrm{d}u=left(1+cos{x}right) mathrm{d}x$

$mathrm{d}v = dfrac{mathrm{d}x}{1+cos(x)}$ $quad rightarrow quad$ $v = int{dfrac{mathrm{d}x}{1+cos{x}}}$

But there is an extra integral to do ( the $v$ function) I evaluated it by sibstitution, and I get $v = tan{dfrac{x}{2}}$, Now

$$int {dfrac{x+sin(x)}{1+cos(x)}}mathrm{d}x = (x+sin{x}), tan{dfrac{x}{2}}-x +C$$
My question: is it possible to evaluate this integral entirely by parts (without using any substitution) ?

I appreciate any ideas

$int dfrac{x + sin x}{1+cos x}dx\
int dfrac{x + 2 sin frac{x}{2} cos frac{x}{2}}{2cos^2 frac{x}{2}} dx\
int frac{1}{2}xsec^2frac{x}{2} + tan frac{x}{2} dx\
int frac{1}{2}xsec^2frac{x}{2}dx + inttan frac{x}{2} dx$

Now we do integration by parts on the $1^{st}$ integral. we won't evaluate the $2^{nd}$ one quite yet.

$u = x, dv = sec^2 frac{x}{2} dx\ du = dx, v = 2 tan frac{x}{2}$

$xtanfrac{x}{2} - int tanfrac{x}{2}+inttanfrac{x}{2} \xtanfrac{x}{2} + C$

You can use the formula:
$$intfrac{f(x)+f'(x)sin x}{1+cos x}dx=frac{f(x)sin x}{1+cos x}+C$$

Because: $$(frac{sin x}{1+cos x})'=frac{1}{1+cos x}$$
then:$$intfrac{f(x)+f'(x)sin x}{1+cos x}dx=int f(x)d(frac{sin x}{1+cos x})+(frac{sin x}{1+cos x})df(x)=int d(frac{f(x)sin x}{1+cos x})=frac{f(x)sin x}{1+cos x}+C$$

Example1:$$int e^x frac{1+sin x}{1+cos x}dx=frac{e^xsin x}{1+cos x}+C$$
Example2:$$int frac{ln x+frac{sin x}{x}}{1+cos x}dx=frac{ln xsin x}{1+cos x}+C$$

You could use $displaystyle v=intfrac{1}{1+cos x}dx=intfrac{1-cos x}{1-cos^2 x} dx=intfrac{1-cos x}{sin^2 x} dx=int(csc^2 x-csc xcot x) dx$

$displaystylehspace{1.05in}=-cot x+csc x=frac{1-cos x}{sin x}=frac{sin x}{1+cos x}$ $;;;$(taking $C=0$)

Write the integrand as

$${x+sin xover1+cos x}={(x+sin x)(1-cos x)over1-cos^2x}=(x+sin x)(1-cos x)csc^2x$$

Now let $u=(x+sin x)(1-cos x)$ and $dv=csc^2x,dx$, for which $v=-cot x$. On noting that

$$du=(1+cos x)(1-cos x)+(x+sin x)(sin x)=xsin x+2sin^2x$$

integration by parts now gives

$$int{x+sin xover1+cos x}dx=(x+sin x)(1-cos x)(-cot x)+int (xcos x+2sin xcos x)dx$$

The integral $int xcos x,dx$ is easily done by parts. The integral $intsin xcos x,dx$ can also, if you like, be done by parts: Letting $u=sin x, dv=cos x,dx$, we get

$$I=intsin xcos x,dx=sin^2x-intcos xsin x,dx=sin^2x-Iimplies I={1over2}sin^2x+C$$