×

# Integral help?

I was doing a problem in which i used a bit unconventional approach, but in my process, i came across the following integral. but i am not really good in integration, as i am new to the subject, so i want to ask, if the integral is derivable? and if so, how to do this:

$\large{\int\dfrac{sin(\frac{nx}{2}).cos\frac{(n+1)x}{2}}{sin(\frac{x}{2})}.dx}$

Note by Aritra Jana
2 years, 1 month ago

Sort by:

Can you post the original problem?? · 2 years, 1 month ago

the original problem asked for the sum of the series :

$$\large{\sum\limits_{r=1}^{n}\frac{1}{r}sin(r\frac{\pi}{3})}$$

I was thinking if i could use the identity :

$$\large{\sum\limits_{r=1}^{n}cos(rx)=\dfrac{sin(\frac{nx}{2}).cos\frac{(n+1)x}{2}}{sin(\frac{x}{2})}}$$

and then integrate this to get the above sum, but as you can see, this was not fruitful.

i shall post this original problem in a new thread. :D · 2 years, 1 month ago

$$\dfrac{\pi}{3} = x$$

$$S = \displaystyle \sum_{r=1}^{n}\dfrac{sin(rx)}{r}$$

$$C = \displaystyle \sum_{r=1}^{n}\dfrac{cos(rx)}{r}$$

$$C + iS = \displaystyle \sum_{r=1}^{n} \dfrac{e^{irx}}{r}$$

$$Now~ \displaystyle \sum_{r=1}^{n} x^{r - 1} = \dfrac{1 - x^n}{1 - x}$$

$$\displaystyle \sum_{r=1}^{n} \dfrac{x^r}{r} = \int \dfrac{1 - x^n}{1 - x}$$

Then keeping $$e^{ix} = x$$ and comparing the imaginary part we can get the answer but I am unable to find this - $$\int \dfrac{x^n}{1 - x}$$ · 2 years, 1 month ago

Any ideas @Jake Lai · 2 years, 1 month ago

There was a problem related to this sum. · 2 years, 1 month ago