# 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
6 years, 3 months ago

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## Comments

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Can you post the original problem??

- 6 years, 3 months ago

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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

- 6 years, 3 months ago

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

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

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

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

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

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

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

- 6 years, 3 months ago

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@U Z Any ideas @Jake Lai

- 6 years, 3 months ago

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There was a problem related to this sum.

- 6 years, 3 months ago

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really? i wasn't aware of that this actually came in our FIITJEE open test Mains

- 6 years, 3 months ago

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