# Evaluation of the integral $\int{\frac{x+\sin{x}}{1+\cos{x}}\mathrm{d}x}$ by parts

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:

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

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

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

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

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

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

$x\tan\frac{x}{2} - \int \tan\frac{x}{2}+\int\tan\frac{x}{2} \\x\tan\frac{x}{2} + C$

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

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

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

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

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

Write the integrand as

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

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

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

integration by parts now gives

$$\int{x+\sin x\over1+\cos x}dx=(x+\sin x)(1-\cos x)(-\cot x)+\int (x\cos x+2\sin x\cos x)dx$$

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

$$I=\int\sin x\cos x\,dx=\sin^2x-\int\cos x\sin x\,dx=\sin^2x-I\implies I={1\over2}\sin^2x+C$$

• thank you , for : $\int{2\cos{x}\sin{x}\,\mathrm{d}x }=\int{\sin{2x}\,\mathrm{d}x } = -\dfrac{\cos{2x}}{2} + C$ – Navaro Apr 9 '16 at 0:20