# Integral of (x + sin(x))/(1 + cos(x))

#### Bart

$$\displaystyle \int{\frac{x+sin(x)}{1+cos(x)}}dx$$

My attempt so far

$$\displaystyle \int{\frac{(x+sin(x))(1-cos(x))}{(1+cos(x))(1-cos(x))}}dx$$ =

$$\displaystyle \int{\frac{x-x*cos(x)+sin(x)-cos(x)sin(x)}{sin^2(x)}}dx$$ =

$$\displaystyle -x*cotan(x)+ln(sin(x))-x*cosec(x)-\frac{cos(x)}{sin^2(x)} + \int{\frac{1}{sin(x)}}dx-ln(sin(x))$$ =

$$\displaystyle -x*cotan(x)-x*cosec(x)-\frac{cos(x)}{sin^2(x)} + \int{\frac{1}{sin(x)}}dx$$

I should get $$\displaystyle \frac{x(1-cos(x))}{sin(x)} = \frac{x}{sin(x)}-x*cotan(x)$$ pretending that I don't know how to solve $$\displaystyle \int\frac{1}{sin(x)}dx$$ so apparantly somehow I should be able to paritally integrate $$\displaystyle \int\frac{x*cos(x)}{sin^2(x)}dx$$ in such a way that I get $$\displaystyle \int\frac{-1}{sin(x)}dx$$ and so get rid of that integral.

Last edited:

#### sa-ri-ga-ma

• Bart and slovakiamaths

#### simplependulum

MHF Hall of Honor
Try to only integrate the first one , there is a trick : we don't need to integrate the second one , it will be deleted finally !!

$$\displaystyle \frac{1}{2} \int x \sec^2(\frac{x}{2})~dx + \int \tan(\frac{x}{2})~dx$$

Using integration by parts ,

$$\displaystyle = x\tan(\frac{x}{2}) - \int \tan(\frac{x}{2})~dx + \int \tan(\frac{x}{2})~dx$$

we can see the second integral can be deleted but note that before writing this

$$\displaystyle = x\tan(\frac{x}{2}) + C$$

we should add one more line , it is

$$\displaystyle = x\tan(\frac{x}{2}) + \int 0~dx$$

so followed by

$$\displaystyle = x\tan(\frac{x}{2}) + C$$

• Bart and slovakiamaths