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:

sin(nx2).cos(n+1)x2sin(x2).dx\large{\int\dfrac{sin(\frac{nx}{2}).cos\frac{(n+1)x}{2}}{sin(\frac{x}{2})}.dx}

Note by Aritra Jana
4 years, 8 months ago

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Can you post the original problem??

Sudeep Salgia - 4 years, 8 months ago

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the original problem asked for the sum of the series :

r=1n1rsin(rπ3)\large{\sum\limits_{r=1}^{n}\frac{1}{r}sin(r\frac{\pi}{3})}

I was thinking if i could use the identity :

r=1ncos(rx)=sin(nx2).cos(n+1)x2sin(x2)\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 - 4 years, 8 months ago

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π3=x\dfrac{\pi}{3} = x

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

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

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

Now r=1nxr1=1xn1x Now~ \displaystyle \sum_{r=1}^{n} x^{r - 1} = \dfrac{1 - x^n}{1 - x}

r=1nxrr=1xn1x \displaystyle \sum_{r=1}^{n} \dfrac{x^r}{r} = \int \dfrac{1 - x^n}{1 - x}

Then keeping eix=x e^{ix} = x and comparing the imaginary part we can get the answer but I am unable to find this - xn1x \int \dfrac{x^n}{1 - x}

U Z - 4 years, 8 months ago

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@U Z Any ideas @Jake Lai

U Z - 4 years, 8 months ago

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There was a problem related to this sum.

Krishna Sharma - 4 years, 8 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 - 4 years, 8 months ago

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