# Integral

• June 20th 2008, 10:58 AM
bobak
Integral
Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak

Edit: the integral is $\int \frac{\sin^2((k+1)x)}{\sin^2x}$
• June 20th 2008, 11:41 AM
Mathstud28
Quote:

Originally Posted by bobak
Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak

Hello,

I am sorry, I hate this notation. Is this

$\sin^2(k+1)\cdot{x}$

Or

$\sin^2((k+1)x)$
?
• June 20th 2008, 11:49 AM
bobak
Quote:

Originally Posted by Mathstud28
Hello,

I am sorry, I hate this notation. Is this

$\sin^2(k+1)\cdot{x}$

Or

$\sin^2((k+1)x)$
?

the second one
• June 20th 2008, 12:03 PM
Moo
Hello :D

Quote:

Originally Posted by bobak
Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak

yay ! I think I found it ! ^^

$\frac{sin[(2k+1)x]}{sin(x)}=\frac{sin(2kx)cos(x)+cos(2kx)sin(x)}{sin (x)}$

$=\frac{2cos(kx)sin(kx)cos(x)+sin(x)[cos^2(kx)-sin^2(kx)]}{sin(x)}$

Multiply above and below by sin(x) :

$=...$ (skipped the step... I guess you can do it (Wink))

Rearranging :

$=\frac{2{\color{red}sin(x)cos(kx)} \cdot {\color{blue}sin(kx)cos(x)}+[{\color{red}sin(x)cos(kx)}]^2-[sin(x)sin(kx)]^2}{sin^2(x)}$

Complete the square in red :

$=\frac{\left(\overbrace{{\color{red}sin(x)cos(kx)} +{\color{blue}sin(kx)cos(x)}}^{sin[(k+1)x]}\right)^2-[{\color{blue}sin(kx)cos(x)}]^2-[sin(x)sin(kx)]^2}{sin^2(x)}$

$\frac{sin[(2k+1)x]}{sin(x)}=\frac{sin^2[(k+1)x]}{sin^2(x)}-\frac{[sin(kx)cos(x)]^2+[sin(x)sin(kx)]^2}{sin^2(x)}$

----------------------------

$\frac{[sin(kx)cos(x)]^2+[sin(x)sin(kx)]^2}{sin^2(x)}$

$=\frac{sin^2(kx)cos^2(x)+sin^2(x)sin^2(kx)}{sin^2( x)}=\frac{sin^2(kx)[cos^2(x)+sin^2(x)]}{sin^2(x)}=\boxed{\frac{sin^2(kx)}{sin^2(x)}}$

----------------------------

Integrating :

-------> $\frac{\pi}{2}=\overbrace{\int_0^{\pi /2} \frac{sin[(2k+1)x]}{sin(x)} \ dx}^{\frac{\pi}{2}}=\int_0^{\pi /2} \frac{sin^2[(k+1)x]}{sin^2(x)} \ dx - \int_0^{\pi /2} \frac{sin^2(kx)}{sin^2(x)} \ dx \quad \square$
• June 20th 2008, 12:20 PM
bobak
Thanks for that Moo. Much apperiaceted.

The last part is pretty straightforward :D

Bobak
• June 20th 2008, 12:22 PM
Moo
Quote:

Originally Posted by bobak
Thanks for that Moo. Much apperiaceted.

The last part is pretty straightforward :D

Bobak

I edited many times my post, so I'm not sure what you're talking about :p
Refresh ^^
• June 20th 2008, 12:26 PM
bobak
Quote:

Originally Posted by Moo
I edited many times my post, so I'm not sure what you're talking about :p
Refresh ^^

Yeah I noticed you editing it, I got the general idea thought, I made the error of trying to work form the result backwards, Which didn't work out very well for me.

Bobak
• June 20th 2008, 03:11 PM
PaulRS
Quote:

Originally Posted by bobak
Hello All

I could do with some help on this integral. I didn't have any problems with part a) and b), I'm stuck on part c).

Bobak

Edit: the integral is $\int \frac{\sin^2((k+1)x)}{\sin^2x}$

In case you want to see an optional solution to the integral in part b) (not much to do with that exercise)

$
I_n = \int_{ - \pi }^\pi {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx} = 2\pi ,\forall n \in \mathbb{N}
$

By Euler's Formula: $
\sin \left( {n \cdot x} \right) = \tfrac{{e^{i \cdot n \cdot x} - e^{ - i \cdot n \cdot x} }}
{{2 \cdot i}};\sin \left( {n \cdot x} \right) = \tfrac{{e^{i \cdot x} - e^{ - i \cdot x} }}
{{2 \cdot i}}
$
(we're not using contours)

Thus $
I_n = \int_{ -\pi }^{\pi } {\frac{{e^{i \cdot
{\left( {2n + 1} \right)}
\cdot x} - e^{ - i \cdot
{\left( {2n + 1} \right)}
\cdot x} }}
{{e^{i \cdot x} - e^{ - i \cdot x} }}dx}
$

Now recall the formula: $
x^n - y^n = \left( {x - y} \right) \cdot \sum\limits_{k = 0}^{n - 1} {x^k \cdot y^{n - 1 - k} }
$
we have: $
e^{i \cdot \left( {2n + 1} \right) \cdot x} - e^{ - i \cdot \left( {2n + 1} \right) \cdot x} = \left( {e^{i \cdot x} - e^{ - i \cdot x} } \right) \cdot \sum\limits_{k = 0}^{2n} {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} }

$

Thus: $
I_n = \int_{ - \pi }^\pi {\left( {\sum\limits_{k = 0}^{2n} {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} } } \right)dx} = \sum\limits_{k = 0}^{2n} {\int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} }

$

Note that: $
\left\{ \begin{gathered}
{\text{If }}k \ne n \Rightarrow \int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} = 0 \hfill \\
{\text{If }}k = n \Rightarrow \int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} = 2\pi \hfill \\
\end{gathered} \right.
$
Therefore: $
I_n = \sum\limits_{k = 0}^{2n} {\int_{ - \pi }^\pi {e^{2 \cdot i \cdot x \cdot \left( {k - n} \right)} dx} } = 2\pi
$
(Wink)

And since: $
\int_{ - \pi }^\pi {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx} = 4 \cdot \int_0^{\tfrac{\pi }
{2}} {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx}
$
we have: $
\int_0^{\tfrac{\pi }
{2}} {\frac{{\sin \left[ {\left( {2n + 1} \right) \cdot x} \right]}}
{{\sin \left( x \right)}}dx} = \tfrac{\pi }
{2},\forall n \in \mathbb{N}
$
• June 20th 2008, 04:33 PM
galactus
I worked out $I_{n}=\int_{0}^{\pi}\frac{sin^{2}(nx)}{sin^{2}(x)} dx$

if you still care.

For all positive integer values of n.

Let's show that $I_{n}=n{\pi}, \;\ n=1,2,3,....$

We have for $n\geq{2}$:

$D_{n}=I_{n}-I_{n-1}=\int_{0}^{\pi}\frac{(sin^{2}(nx)-sin^{2}((n-1)x)}{sin^{2}(x)}dx$

$=\int_{0}^{\pi}\frac{(sin(nx)-sin((n-1)x))(sin(nx)+sin((n-1)x))}{sin^{2}(x)}dx$

$\int_{0}^{\pi}\frac{2sin(\frac{x}{2})cos(n-\frac{x}{2})2sin(nx-\frac{x}{2})cos(\frac{x}{2})}{sin^{2}(x)}dx$

$=\int_{0}^{\pi}\frac{sin(x)sin((2n-1)x)}{sin^{2}(x)}dx$

$=\int_{0}^{\pi}\frac{sin((2n-1)x)}{sin(x)}dx$

So, $D_{n}=J_{2n-1}, \;\ n\geq{2}$

$J_{m}=\int_{0}^{\pi}\frac{sin(mx)}{sin(x)}dx, \;\ m=0,1,2,..$

Now, for $m\geq{2}$:

$J_{m}-J_{m-2}=\int_{0}^{\pi}\frac{sin(mx)-sin((m-2)x)}{sin(x)}dx$

$=\int_{0}^{\pi}\frac{2sin(x)cos((m-1)x)}{sin(x)}dx$

$=2\int_{0}^{\pi}cos((m-1)x)dx$

$=2\left[\frac{sin((m-1)x)}{m-1}\right]_{0}^{\pi}=0$

So, we have

$J_{m}=J_{m-2}=J_{m-4}=....=\begin{gathered}J_{0}=0, \;\ \text{if m even}\\ J_{1}={\pi}, \;\ \text{if m odd}\end{gathered}$

Therefore, hence and hereftofore:

We get $D_{n}={\pi}, \;\ n\geq{2},$ so that $I_{n}=n{\pi}, \;\ n\geq{1}, \;\ as \;\ I_{1}={\pi}$

WHEW!!

I just noticed. I used Pi instead of Pi/2 all this time. Oh well. Alas.
• June 21st 2008, 01:13 AM
Moo
Hi galactus... Isn't it too complicated ? :eek:

$\int_0^{\pi /2} \frac{\sin^2 (kx)}{\sin^2(x)} \ dx=k \frac{\pi}{2}$

---------------------
For k=0, $\frac{\sin^2(kx)}{\sin^2(x)}=0 \quad \square.$

Assume that $\int_0^{\pi /2} \frac{\sin^2(kx)}{\sin^2(x)} \ dx = k \frac{\pi}{2}$

From the previous question :

$\int_0^{\pi /2} \frac{\sin^2 [(k+1)x]}{\sin^2(x)} \ dx=\int_0^{\pi /2} \frac{\sin^2 (kx)}{\sin^2(x)} \ dx + \frac{\pi}{2}=\bf (k+1) \frac{\pi}{2} \quad \blacksquare.$

huh?