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Given integral:$$I=\int\,\,{{\sin3x}\cdot\sin{5x\over2}\over\sin {x\over2}}\,dx$$

I have solved the given integral but it's too lengthy and tiresome to write here.Also is there any method to solve the integral with help of Complex Integration

(I'm new here on math.SE and can't type frequently so for me it'll take too much time to type the complete solution here since I'm using iPad)

In case anyone wants to see my solution, I'll attach the image of paper-solution here Sorry for not being supportive, need your help!

Michael Rozenberg
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Arpit Yadav
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  • Limits of integration could make this way prettier. Just to confirm, you are looking for the antiderivative? – Brevan Ellefsen Jul 25 '17 at 01:58
  • @BrevanEllefsen It's an indefinite integral.The limit of Integration isn't given.thanks. – Arpit Yadav Jul 25 '17 at 02:00
  • @BrevanEllefsen, I didn't go with substitution, i used trigonometric identities but using that makes problem too lengthy. – Arpit Yadav Jul 25 '17 at 02:15
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    Ok, better idea. Just use the sub $t = x/2$ and the fact that $\sin(5t) = \sin(t)(2\cos \left(2t\right)+2\cos \left(4t\right)+1)$ – Brevan Ellefsen Jul 25 '17 at 02:17
  • @BrevanEllefsen I had multiplied $\cos {5x\over2}$ in Numerator and Denominator and then proceed. – Arpit Yadav Jul 25 '17 at 02:17
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    Depends on what math you know. I would just note that $\cos(5t)+i\sin(5t) = \left(\cos(t)+i\sin(t)\right)^5$ and expand the RHS and take the imaginary part! This obviously generalizes, and so the formulas for $\sin(ax)$ and $\cos(ax)$ are simple results of the Binomial Formula. – Brevan Ellefsen Jul 25 '17 at 02:28
  • @BrevanEllefsen: If you have something that you think is an answer, please add an answer! Please see: https://math.meta.stackexchange.com/questions/1559/dealing-with-answers-in-comments – Aryabhata Jul 25 '17 at 02:30
  • @BrevanEllefsen, I'm used $\sin{5t}=\sin{(4t+t)}$ and the expanded it. – Arpit Yadav Jul 25 '17 at 02:33
  • @Aryabhata oh, my comment was more of a hint. Your post is more of an answer. I don't see any reason why both can't coexist :) I just didn't feel like this question was unique and applicable to a large enough audience for me to expand my comments into a full answer – Brevan Ellefsen Jul 25 '17 at 02:33
  • @ArpitYadav Sweet, that is also a completely valid proof. It might leave things in a little messier of a form than I give above, but it shouldn't matter if you do a little work, – Brevan Ellefsen Jul 25 '17 at 02:34
  • @BrevanEllefsen What to do if the given problem is generalized or for given with bigger no., then how does expansion works? – Arpit Yadav Jul 25 '17 at 02:50
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    @ArpitYadav what kind of generalization do you have in mind? There are infinitely many generalizations :) I would imagine you are asking either how to integrate $\frac{\sin(ax)\sin(bx)}{\sin(cx)}$, how to integrate $\frac{\sin(ax)\sin(bx)}{\sin(x)}$, how to integrate $\frac{\sin(ax)\sin((a+1)x)}{\sin(cx)}$ or how to integrate $\frac{\sin(ax)\sin((a+1)x)}{\sin(x)}$, but the best path to take depends on what you have in mind. The first form I give is probably the most general of the simple generalizations of your problem, though I'm not sure how clean an answer there is – Brevan Ellefsen Jul 25 '17 at 03:00
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    Tests show that the first the form above yields results in terms of the sum of logarithms of simple trig functions for any values of $a,b,c$ I throw in. I will see if Mathematica can find a general form – Brevan Ellefsen Jul 25 '17 at 03:04
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    @ArpitYadav: A more general approach: https://math.stackexchange.com/questions/29980/evaluating-int-p-sin-x-cos-x-textdx – Aryabhata Jul 25 '17 at 03:33
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    While I am not sure if a pretty closed form exists, we can definitely integrate $\frac{\sin(ax)\cos(bx)}{\sin(cx)}$ in the case $c=1$ by noting $\frac{\sin(ax)\cos(bx)}{\sin(x)} = \frac{1}{2}\frac{\cos(px)}{\sin(x)}-\frac{1}{2}\frac{\cos(qx)}{\sin(x)}$ where $p=a-b$ and $q=a+b$ by expanding the numerators using binomial expansion, which just varies a bit depending on whether $p$ is odd is even. This also handles any cases where $a$ or $b$ are multiples of $c$, or even $p>c$ if we are careful I think. As to $p<c$, I would have to play around with the function more. I hope this answer suffices! – Brevan Ellefsen Jul 25 '17 at 03:42
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    This generally settles the question when $\min{a-b, a+b} > c$. However, when this condition does not hold things can get a lot worse than sums of logarithms of basic trig functions and constants... in general I can only find closed forms when $c$ is even, and I don't even know if that continues. Odd values of $c$ seem to depend on the roots of polynomials with increasing degree $d$ as $c$ increases (for small values of $c$, it seems that $d$ increases by $1$ each time $c$ reaches a new prime) which means that, unless we have really nice polynomials, the constants aren't elementary – Brevan Ellefsen Jul 25 '17 at 04:13

4 Answers4

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$$\frac{\sin3x\sin\frac{5}{2}x}{\sin{\frac{x}{2}}}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}(3-4\sin^2x)\sin\frac{5}{2}x}{\sin{\frac{x}{2}}}=$$ $$=(\sin2x+\sin3x)(1+2\cos2x)=\sin2x+\sin3x+\sin4x+\sin5x+\sin{x}$$

Michael Rozenberg
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  • Ah neat! Now I know why it looked so familar: https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro. See also: Dirichlet Kernel. – Aryabhata Jul 25 '17 at 17:09
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Using $\sin\theta=\frac{e^{i\theta}-e^{i\theta}}{2i}$, one has \begin{eqnarray} &&{{\sin3x}\cdot\sin{5x\over2}\over\sin {x\over2}}\\ &=&\frac{1}{2i}{{(e^{3ix}-e^{-3ix})}\cdot(e^{5ix/2}-e^{-5ix/2})\over e^{ix/2}-e^{-ix/2}}\\ &=&\frac{1}{2i}{{(e^{3ix}-e^{-3ix})}\cdot(e^{3ix}-e^{-2ix})\over e^{ix}-1}\\ &=&\frac{1}{2ie^{5ix}}{{(e^{6ix}-1)}(e^{5ix}-1)\over e^{ix}-1}\\ &=&\frac{1}{2ie^{5ix}}(e^{6ix}-1)(e^{4ix}+e^{3ix}+e^{2ix}+e^{ix}+1)\\ &=&\frac{1}{2ie^{5ix}}(e^{10ix}+e^{9ix}+e^{8ix}+e^{7ix}+e^{6ix}-e^{4ix}-e^{3ix}-e^{2ix}-e^{ix}-1)\\ &=&\frac{1}{2i}(e^{5ix}+e^{4ix}+e^{3ix}+e^{2ix}+e^{ix}-e^{-ix}-e^{-2ix}-e^{-3ix}-e^{-4ix}-e^{-5ix})\\ &=&\sin(5x)+\sin(4x)+\sin(3x)+\sin(2x)+\sin x. \end{eqnarray}

xpaul
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Set $y = x/2$ and then use the identity

$$\sin(5y) = \sin y(2\cos 2y + 2 \cos 4y + 1)$$

Then you will essentially have to compute

$$ \int \sin(6y)(2\cos 2y + 2 \cos 4y + 1) \text{d}y$$

Now you can use the identity $$2 \sin a \cos b = \sin(a+b) + \sin(a-b)$$

The rest if left as an exercide.

For completeness: to prove the identity:

$$\sin(5y) = \sin y(2\cos 2y + 2 \cos 4y + 1)$$

Consider

$$\sin 5x - \sin x = 2 \cos 3x \sin 2x$$

(using $\sin (a+b) - \sin (a-b) = 2 \cos a \sin b$)

$$ 2 \cos 3x \sin 2x = 4 \cos 3x \cos x \sin x$$

(using $\sin 2x = 2 \sin x \cos x$)

$$ = 2\sin x (\cos 4x + \cos 2x)$$

Using $\cos (a+b) + \cos (a-b) = 2 \cos a \cos b$

Aryabhata
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Hint:

If $f(m, n)=\dfrac{\sin mx\sin nx}{\sin\dfrac x2}, $

$f(m, n)-f(m, n-1)=2\sin mx\cos\dfrac{(2n-1)x}2=? $( Use Werner's formula)

$$\implies\int f(m,n)dx=\int f(m,n-1)dx+\int\left(\sin\dfrac{2m+2n-1}2+\sin\dfrac{2m-2n+1}2\right)dx$$

Set $m=\dfrac52,\dfrac32$ and add

Now $\displaystyle\int f\left(m,\dfrac12\right)dx=?$

Finally set $m=3$

lab bhattacharjee
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