2
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ Ok, better idea. Just use the sub $t = x/2$ and the fact that $\sin(5t) = \sin(t)(2\cos \left(2t\right)+2\cos \left(4t\right)+1)$ $\endgroup$ – Brevan Ellefsen Jul 25 '17 at 2:17
  • 1
    $\begingroup$ 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. $\endgroup$ – Brevan Ellefsen Jul 25 '17 at 2:28
  • 1
    $\begingroup$ @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 $\endgroup$ – Brevan Ellefsen Jul 25 '17 at 3:00
  • 1
    $\begingroup$ 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 $\endgroup$ – Brevan Ellefsen Jul 25 '17 at 3:04
  • 1
    $\begingroup$ @ArpitYadav: A more general approach: math.stackexchange.com/questions/29980/… $\endgroup$ – Aryabhata Jul 25 '17 at 3:33
7
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ help me to understand the 4th step where you write $${e^{5ix}-1}\over{e^{ix}-1}$$ as $$e^{4ix}+e^{3ix}+e^{2ix}+e^{ix}+1$$ $\endgroup$ – Arpit Yadav Jul 26 '17 at 12:12
  • 1
    $\begingroup$ @ArpitYadav, use $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$. $\endgroup$ – xpaul Jul 26 '17 at 13:28
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Precisely what I intended in my comment above - thanks for writing it out. $\endgroup$ – Brevan Ellefsen Jul 25 '17 at 2:24
  • $\begingroup$ @Aryabhata, Brevan Ellefsen already mentioned this method in comments,but Thanks for the time. $\endgroup$ – Arpit Yadav Jul 25 '17 at 2:25
  • $\begingroup$ @ArpitYadav: I didn't see that comment. Interesting. In any case, it is better to leave answers than comments (google search does not pick up comments for instance) and this answer is intended not just for you but for any future readers that come across this. $\endgroup$ – Aryabhata Jul 25 '17 at 2:29
  • $\begingroup$ @Aryabhata, Thanks for the time though. $\endgroup$ – Arpit Yadav Jul 25 '17 at 2:40
  • $\begingroup$ @ArpitYadav: No worries. Had some free time :) $\endgroup$ – Aryabhata Jul 25 '17 at 2:42
1
$\begingroup$

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 $m=3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.