find integral $\int_0^{\pi/2}~{\frac{\sin(~40~x)}{\sin(~5~x)}}\mathrm{d}x$

Question

Find integral : $$\int_0^{\frac{\pi}{2}}~{\frac{\sin(~40~x)}{\sin(~5~x)}~dx}$$

My attempt -writing $$\sin~40x= \sin~40x - \sin~30x + \sin~30x - \sin~20x + \sin~20x - \sin~10x +\sin~10x$$

and apply formula $\sin~C - sin~D$ and write $\sin~10x = 2~\sin~5x\cos~5x$ … by this $\sin~5x$ will be eliminated and will be left with only cos terms. Any other approach for this problem.

Answer 1

Hint: Note that $$\frac {\sin 40x-\sin 10x}{\sin 5x}=2(\cos 15x+\cos 25x +\cos 35x)$$

Answer 2

Expand the numerator, then divide each term by $\sin(5x)$ (or $\sin(y)$ after substituting $y=5x$):

$$\begin{align} I&=\int_0^{\frac\pi2}\frac{\sin(40x)}{\sin(5x)}\,\mathrm dx\\[1ex] &= \frac15\int_0^{\frac{5\pi}2}\frac{\sin(8y)}{\sin(y)}\,\mathrm dy\\[1ex] &=\frac15\int_0^{\frac{5\pi}2}\frac{f(y)}{\sin(y)}\,\mathrm dy \end{align}$$

where

$$\begin{align}f(y)&=\operatorname{Im}(e^{8iy})\\&=8\cos^7(y)\sin(y)-56\cos^5(y)\sin^3(y)+56\cos^3(y)\sin^5(y)-8\cos(y)\sin^7(y)\end{align}$$

$$\implies I = \frac15\int_0^{\frac{5\pi}2}\bigg(8\cos^7(y)-56\cos^5(y)\sin^2(y)+56\cos^3(y)\sin^4(y)-8\cos(y)\sin^6(y)\bigg)\,\mathrm dy$$

Answer 3

You can generalize to any integer:

$\displaystyle\frac{\sin(nx)}{\sin(x)}=\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}=e^{-i(n-1)x}\left(\frac{1-e^{i2nx}}{1-e^{i2x}}\right)=e^{-i(n-1)x}\sum\limits_{k=0}^{n-1}e^{i2kx}=\sum\limits_{k=0}^{n-1}e^{i(2k+1-n)x}$

From there you have two possibilities, notice the indexes $(2k+1-n)$ take values

• $\begin{cases}-(n-1),-(n-3),\cdots,-1,1,\cdots,(n-3),(n-1)&n\text{ even}\\-(n-1),-(n-3),\cdots,-2,0,2,\cdots,(n-3),(n-1)&n\text{ odd}\end{cases}$

Therefore you can regroup terms two by two $\, e^{imx}+e^{-imx}=2\cos(mx)$ and end up with a sum of $\cos(mx)$ where $m$ varies with a step $2$.

These cosinus are easy to integrate and will give you a rational value for the integral. Notice that for $n$ odd, you have also a constant term that will make a $\dfrac{\pi}2$ value appear.

• The second possibility is to integrate the exponential directly

Then $\displaystyle I_n=\sum\limits_{k=0}^{n-1}\dfrac{e^{\frac{i\pi}2(2k+1-n)}-1}{(2k+1-n)i}$

Notice also that $\displaystyle \int_0^{5\pi/2}\frac{\sin(8x)}{\sin(x)}dx=\int_0^{\pi/2}\frac{\sin(8x)}{\sin(x)}dx\ $ since the integral on $[0,2\pi]$ is zero.

So overall our integral is just $\dfrac 15I_8$

Answer 4

Hint:

Set $5x=y$

Using $\sin2A=\sin A\cos A,$

$$\sin8y=\cdots=8\sin y\cos y\cos2y\cos4y$$

If $\sin y\ne0,$

$$\dfrac{\sin8y}{\sin y}=8\cos y\cos2y\cos4y$$

$$=4\cos y(2\cos2y\cos4y)$$

$$=4\cos y(\cos2y+\cos6y)$$

$$=2(\cos y+\cos3y+\cos5y+\cos7y)$$

Generalization:

Using $\sum \cos$ when angles are in arithmetic progression,

$$2\sin B\sum_{r=0}^{n-1}\cos(2r+1)A=\sin2nA$$ Here $n=4$