4
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

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

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

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $C=0$)

$\endgroup$
1
$\begingroup$

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.