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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, 3 months ago

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Can you post the original problem?? Sudeep Salgia · 2 years, 3 months ago

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@Sudeep Salgia 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 Aritra Jana · 2 years, 3 months ago

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@Aritra Jana \(\dfrac{\pi}{3} = x\)

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

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

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

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

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

Then keeping \( e^{ix} = x\) and comparing the imaginary part we can get the answer but I am unable to find this - \( \int \dfrac{x^n}{1 - x}\) Megh Choksi · 2 years, 3 months ago

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@Megh Choksi Any ideas @Jake Lai Megh Choksi · 2 years, 3 months ago

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@Aritra Jana There was a problem related to this sum. Krishna Sharma · 2 years, 3 months ago

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@Krishna Sharma really? i wasn't aware of that this actually came in our FIITJEE open test Mains Aritra Jana · 2 years, 3 months ago

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